Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

Question 1.
Find the rank of the matrix
A = \(\left(\begin{array}{cccc}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
Solution:
\(\left(\begin{array}{cccc}
1 & -3 & 4 & 7 \\
9 & 1 & 2 & 0
\end{array}\right)\)
The order of A is 2 × 4
∴ P(A) ≤ 2
Let us transform the matrix A to an echelon form

The number of non-zero rows is 2
∴ P(A) = 2

Question 2.
Find the rank of the matrix
A = \( \left[\begin{array}{cccc}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right]\)
Solution:
A = \( \left[\begin{array}{cccc}
-2 & 1 & 3 & 4 \\
0 & 1 & 1 & 2 \\
1 & 3 & 4 & 7
\end{array}\right]\)
The order of A is 3 × 4
∴ P(A) < 3
Let us transform the matrix A to an echelon form

The number of non-zero rows = 3
∴ P(A) = 3

Question 3.
Find the rank of the matrix
A = \(\left[\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{llll}
4 & 5 & 2 & 2 \\
3 & 2 & 1 & 6 \\
4 & 4 & 8 & 0
\end{array}\right]\)
The order of A is 3 × 4
∴ p(A) < 3
Let us transform the matrix A to an echelon form


The last equivalent matrix is in the echelon form.
Number of non-zero rows = 3
∴ P(A) = 3

Question 4.
Examine the consistency of the system of equations;
x + y + z = 7, x + 2y + 3z = 18, y + 2z = 6
Solution:
x + y + z = 7
x + 2y + 3z = 18
y + 2z = 6
The matrix form of the equations

Here p(A) ≠ p(A, B)
∴ The given system is inconsistent and has no solution.

Question 5.
Find k if the equations 2x + 3y – z = 5, 3x – y + 4z = 2, x + 7y – 6z = k are consistent.
Solution:
2x + 3y – z = 5
3x – y + 4z = 2
x + 7y – 6z = k
The matrix form of these equations

Obviously, the last equivalent matrix is in the ech-elon form.
Since the equations are consistent
P(-A) = p(A, B)
p(A) = 2 and p (A, B) = 2 then
k = 8

Question 6.
Find k if the equations x + y + z = 1, 3x – y – z = 4, x + 5y + 5z = k are inconsistent.
Solution:
x + y + z = 1
3x – y – z = 4
x + 5y + 5z = k
The matrix equation corresponding to the given system is

Obviously, the last equivalent matrix is in the ech-elon form.
Since the equations are inconsistent
p(A) ≠ p (A, B)
Here p(A) = 2 but p(A, B) should not equal to 2
∴ k ≠ 0
The equations are inconsistent when k assume any real value other than 0.

Question 7.
Solve the equations x + 2y + z = 7, 2x – y + 2z = 4, x + y – 2z = -1 by using Cramer’s rule.
Solution:
x + 2y + z = 7
2x – y + 2z = 4
x + y – 2z = -1
Here Δ = \(\left|\begin{array}{ccc}
1 & 2 & 1 \\
2 & -1 & 2 \\
1 & 1 & -2
\end{array}\right|\)
= 1(2 – 2)-2 (-4 – 2) + 1(2 + 1)
= 1(0) -2 (-6) + 1(3)
= 12 + 3 = 15 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
7 & 2 & 1 \\
4 & -1 & 2 \\
-1 & 1 & -2
\end{array}\right|\)
= 7 (2 – 2) -2 (-8 + 2) + 1(4 – 1)
= 7(0)-2 (-6) + 1(3)
= 12 + 3
= 15
Δ_y = \(\left|\begin{array}{ccc}
1 & 7 & 1 \\
2 & 4 & 2 \\
1 & -1 & -2
\end{array}\right|\)
= 1(-8 + 2) -7 (-4 – 2) + 1(-2 – 4)
= 1 (-6) – 7 (-6) + 1 (-6)
= -6 + 42 – 6 = 30
Δ_z = \(\left|\begin{array}{ccc}
1 & 2 & 7 \\
2 & -1 & 4 \\
1 & 1 & -1
\end{array}\right|\)
= 1(1 – 4) -2 (-2 – 4) + 7 (2 + 1)
= 1 (-3) -2 (-6) + 7 (3)
= -3 + 12 + 21 = 30
∴ By cramer’s rule

∴ The Solution is (x, y, z) = (1, 2, 2)

Question 8.
The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is Rs 220. Find the cost of each per kg. using Cramer’s rule.
Solution:
Let cost of 1 kg of wheat be x; cost of 1kg of sugar be y and cost of 1 kg of rice be z.
2x + y = 100
x + z = 80
3x + 2y + z = 220
Here Δ = \(\left|\begin{array}{ccc}
2 & 1 & 0 \\
1 & 0 & 1 \\
3 & 2 & 1
\end{array}\right|\)
= 2 (0 – 2) -1 (1 – 3)
= – 4 + 2 = -2
≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
100 & 1 & 0 \\
80 & 0 & 1 \\
220 & 2 & 1
\end{array}\right|\)
= 100 (0 – 2) -1 (80 – 220)
= – 200 -1 (-140)
= – 200 +140
= -60
Δ_y = \(\left|\begin{array}{ccc}
2 & 100 & 0 \\
1 & 80 & 0 \\
3 & 220 & 1
\end{array}\right|\)
= 2 (80 – 220) -100 (1 – 3)
= 2 (-140) – 100 (-2)
= – 280 + 200
= -80
Δ_x = \(\left|\begin{array}{ccc}
2 & 1 & 100 \\
1 & 0 & 80 \\
3 & 2 & 220
\end{array}\right|\)
= 2 (0 -160) -1 (220 – 240) + 100 (2 – 0)
= 2 (-160) -1 (-20) + 100 (2)
= -320 + 20 + 200
= -100
∴ By cramer’s rule

The solution is (x, y, z) = (30, 40, 50)
∴ The cost of 1 kg of wheat is Rs 30; the cost of 1 kg of sugar is Rs 40 and the cost of 1 kg of rice is Rs 50.

Question 9.
A salesman has the following record of sales during three items A, B, and C, which has different rates of commission.

Find out the rate of commission on items A, B, and C by using Cramer’s rule.
Solution:
Let x, y, and z be the commission on items A, B, and C respectively.
90x + 100y + 20z = 800 ⇒ 9x + 10y + 2z = 80 ……… (1)
130x + 50y + 40z = 900 ⇒ 13x + 5y + 4z = 90 ………. (2)
60x + 100y + 30z = 850 ⇒ 6x + 10y + 3z = 85 ………. (3)
Here Δ = \(\left|\begin{array}{ccc}
9 & 10 & 2 \\
13 & 5 & 4 \\
6 & 10 & 3
\end{array}\right|\)
= 9 (15 – 40) -10 (39 – 24) + 2 (130 – 30)
= 9 (-25) – 10 (15) + 2 (100)
= -225 – 150 + 200
= -175
≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
80 & 10 & 2 \\
90 & 5 & 4 \\
85 & 10 & 3
\end{array}\right|\)
= 80 (15 – 40) -10 (270 – 340) + 2 (900 – 425)
= 80 (-25) -10 (-70) + 2 (475)
= -2000 + 700 + 950
= -350
Δ_y = \(\left|\begin{array}{ccc}
9 & 80 & 2 \\
13 & 90 & 4 \\
6 & 85 & 3
\end{array}\right|\)
= 9 (270 – 340) – 80 (39 – 24) + 2 (1105 – 540)
= 9 (-70)-80 (15) + 2 (565)
= -630 – 1200 + 1130
= -700
Δ_z = \(\left|\begin{array}{ccc}
9 & 10 & 80 \\
13 & 5 & 90 \\
6 & 10 & 85
\end{array}\right|\)
= 9 (425 – 900) -10 (1105 – 540) + 80 (130 – 30)
= 9 (-475) -10 (565) + 80 (100)
= – 4275 – 5650 + 8000
= -1925
∴ By cramer’s rule

The solution is (x, y, z) = (2, 4, 11)
∴ The rates of commission for A, B and C are Rs 2, Rs 4 and Rs 11 respectively.

Question 10.
The subscription department of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 60% of those who already subscribe will subscribe again while 25% of those who do not now subscribe will subscribe. In the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
The transition probability matrix is S

∴ 39% of those receiving the current letter can be expected to order a subscription.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 15 Recent Trends in Marketing Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions 15 Recent Trends in Marketing

12th Commerce Guide Recent Trends in Marketing Mix Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
Selling goods/ services through internet is
a) Green marketing
b) E- business
c) Social marketing
d) Meta marketing
Answer:
b) E-business

Question 2.
Which is gateway to internet?
a) Portal
b) CPU
c) Modem
d) Webnaire
Answer:
c) Modem

Question 3.
Social marketing deals with:
a) Society
b) Social Class
c) Social change
d) Social evil
Answer :
c) Social change

Question 4.
Effective use of Social media marketing increase conversion rates of –
a) Customer to buyers
b) Retailer to customers
c) One buyer to another buyers
d) Direct contact of marketer
Answer:
a) Customer to buyers

Question 5.
A company’s products and prices are visually represented by
a) Shopping cart
b) Web portal
c) Electronic catalogue
d) Revenue model
Answer:
c) Electronic catalogue

Question 6.
Pure play retailers are called
a) Market creators
b) Transaction brokers
c) Merchants
d) Agents
Answer:
b) Transaction brokers

II. Very Short Answer Questions.

Question 1.
What is E-Business?
Answer:
If all the business transactions are carried out through the internet and other online tools, it is called E-business.

Question 2.
What is green marketing?
Answer:

• It refers to a holistic marketing concept with growing awareness about the implications of Global warming, non -biodegradable solid waste, the harmful impact of pollutants, etc.
• It is also known as “Environmental marketing”, “Ecological marketing” Eco-friendly marketing and sustainable marketing”.

Question 3.
What is service marketing?
Answer:
Service marketing denotes the processing of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism, professional services, repairs etc.

Question 4.
Define E-Marketing.
Answer:
“E-Marketing refers to the application of a broad range of information technology for creating more customer value through more effective segmentation, targeting differentiation and positioning strategies, planning more efficiently and executing the conception distribution, promotion and pricing of goods, services and ideas and creating exchanges that satisfy individual consumer and organization’s consumer objectives”. _ JUDY STRAUSS

Question 5.
What is E-Tailing?
Answer:
E-tailing or electronic retailing refers to the selling of goods and services through a shopping website (internet) or through a virtual Store to the ultimate consumer.

Question 6.
What is social marketing?
Answer:

• Social marketing is a new marketing tool
• It is the systematic application of marketing philosophy and techniques to achieve specific behavioural goals that ensure social good
• [Eg] Asking people- Not to smoke in a public place.

III. Short Answer Questions.

Question 1.
What are the advantages of E-Marketing?
Answer:

• Any time market: E-Marketing provides 24 hours and 7 days “24/7” Service to its users. So consumer can shop or order the product anytime from anywhere.
• Direct contact of end consumer by the manufacturer cuts down the substantially intermediation cost. Thus products bought through e-marketing become cheaper.
• Customers can buy whatever they want / need just by browsing the various sites.

Question 2.
Discuss the objectives E-Marketing (EARS)
Answer:

• Expansion of market share.
• Achieving higher brand awareness.
• Reduction of distribution and promotional expenses
• Strengthening database.

Question 3.
Explain in detail Niche marketing. [Nitch] [Neesh]
Answer:

• “Niche marketing “ denotes a strategy of directing all marketing efforts towards one well-defined segment of the population.
•  Actually, there is no market in a niche market.
•  It is found by the company, by identifying the need of customers who are not served or underserved by the competitors.
• It involves a specific target audience with a specialized offering.
•  It aimed at being a big fish in a small pond instead of being a small fish in a big pond.
•  [eg] Not everybody wants to watch the movie by paying 5x-6xtimes the cost of the normal tick.

IV. Long Answer Questions.

Question 1.
Explain in detail how traditional marketing differ from E-marketing [FEETS].
Answer:

Question 2.
Explain the advantages and disadvantages of E-marketing.
Answer:
E-tailing or electronic retailing refers to the selling of goods and services through a shopping website:
Advantages:

1. Customer can buy the product at any time from anywhere.
2. Direct contact of end consumer by the manufacturer cuts down the cost.
3. Customer can buy whatever they want by browsing the various sites.

Disadvantages:

1. E-tailing needs a strong advertisement and for which it has to spend large amount.
2. It is not suitable for small size business.

Question 3.
Discuss any two new methods of marketing with their advantages.
Answer:|
1. Rural Marketing:
Rural marketing is a process of developing pricing, promoting, and distributing rural specific goods and services leading to exchange with rural customers. There is an inflow of goods into rural markets for production and consumption and there is also an outflow of products to urban areas. the rural to urban flow consists of agricultural products like rice, wheat and sugar, etc.

2. Service Marketing:
Service marketing is a special branch of marketing. It denotes the processing of selling service goods like telecommunication, banking, insurance, healthcare, tourism, and professional services. The service products are mostly intangible. The unique feature of services marketing warrants different strategies compared with the marketing of physical goods.

12th Commerce Guide Recent Trends in Marketing Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
Pick the odd one out:
a) E-Commerce
b) E-Marketing
c) E-Tailing
d) E-Banking
Answer:
d) E-Banking

Question 2.
Find out which is not suitable? The products marketed in commodity exchange are ______
(a) Crude oil
(b) Rice
(c) Copper
(d) Gold
Answer:
(b) Rice

Question 3.
…………… is marketing knowns as environmental marketing or Ecological marketing or Eco-friendly marketing or sustainable marketing
a) Green
b) Social
c) Service
d) Rural
Answer:
a) Green

Question 4.
…………….. Marketing is holistic with growing awareness about implications of Global warming Non-Bio degradable solid waste harmful impact of pollutants etc.
a) Rural
b) Traditional
c) Service
d) Green
Answer:
d) Green

Question 5.
Which one of the following is not correctly matched?
a) Referral marketing – Reference
b) Content marketing – Communicating
c) E-Tailing – Online
d) Service marketing – Anti-drug
Answer:
d) Service marketing – Anti-drug

Question 6.
Select the odd one out.
a) Electronic marketing
b) Online marketing
c) Web marketing
d) Traditional marketing
Answer:
d) Traditional Marketing

Question 7.
Which one represents a cluster of manufacturers, content providers and online retailers organised around an activity?
a) Virtual mall
b) Association
c) Metomediary
d) Portal
Answer:
c) Metomediary

Question 8.
Green Shelter concept was introduced by group:
a) ACME
b) Tata
c) Reliance
d) ICI
Answer:
a) ACME

II. Match The Following

1. Match list -1 with list 2

Answer:
a) (i) 4, (ii) 3, (iii) 1, (iv) 2

Question 2.

Answer:
c) (i) 2,(ii) 3,(iii) 1,(iv) 4

Question 3.

Answer:
a) (i) 4, (ii) 3, (iii) 2, (iv) 1

III. Assertion and Reason

Question 1.
Assertion [ A] : Actually there is no market in niche market
Reason [B]: It is found by identifying the need of customers which are not served and under served by the competitions .
a) A is true R is false
b) A is false R is true
c) A and R are true
d) A and R are false
Answer:
c) A and R are true

Question 2.
Assertion [A]: “Offer more – for less “ is a strategy to satisfy the consumers
Reason [R ]: They use computers, laptops tablets smart-phone android phone to access different websites.
a. A is true R is false
b. A and R are true
c. A and R are false
d. A and R are false
Answer:
b) A and R are true

IV. Short Answer Questions.

Question 1.
What is B2B and B2C type of E-Commerce?
Answer:
B2B-Business to Business:

• Under this model .Business concerns transact with one another through internet.
• [eg] Flipkart

B2C- Business to consumer:

• Under this model Business concerns sell directly to the consumers.
• [eg] Shopping zone .

Question 2.
Explain the importance of social marketing ?
Answer:

• It helps to eradicate social evils that affect the society and quality of life
• It promotes the consumption of socially desirable products and develops health consciousness.
• It is a new marketing tool by using philosophy and techniques to achieve specific behavioural goals which ensure social goal.

Question 3.
Elucidate how E-Commerce differs from E-Business.
Answer:

• E-Commerce simply refers to the buying and selling of products and services through online.
• E-Business goes away beyond the simple buying and selling goods and services and much wider range of business processes.
• E-Commerce and E-Business is used interchangeably in its broader meaning just as trade and commerce.

V. Long Answer Questions.

Question 1.
Describe the various strategies pursued in recent day’s marketers.
Answer:
Viral Marketing:

• “Viral marketing “ is a marketing technique that impels the users to pass on a marketing message to other users, creating a potentially exponential in the message visibility and effect.
•  It is able to generate interest and potential sale of a brand or product through messages that
  spread quickly like a virus from person to person.
• [eg] Emergency needs through you tube, face book, twitter etc… [Need of blood -to save the life]

Guerrilla marketing:
“ Guerilla marketing “ represents an advertisement strategy to promote or services on streets or other public places with monkey like malls, parks, beach etc.

Referral marketing:
“Referral marketing is the method of promoting products or services to new customers through existing customers.

Ambush marketing:

• It is a new technique where by a particular advertiser seeks to connect his product to the event in the mind of polentional customers without paying sponsoring expenses to the event.
• [eg] X has sponsored a cricket match to promote his brand. A group of people sitting on the gallery wearing Y’ brand name T- shirts come in to focuses, in this case ‘Y’ promotes his brand at ‘X’ expense.

Multilevel marketing:

• It is the marketing strategy where in the direct sales companies encourage its existing distributors
• ID recruit new distributors to facilitate the sale of goods and services.

Content marketing:
It is said to be the art of communicating with customers and prospects without selling.

Question 2.
Compare the concept of social marketing with service marketing. (BEEP)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Choose the most suitable answer from the given four alternatives:

Question 1.
A = (1, 2, 3), then the rank of AA^T is
(a) 0
(b) 2
(c) 3
(d) 1
Solution:
(d) 1
Hint:
A = (1, 2, 3) then
A^T = \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\)
AA^T = (1, 2, 3) \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\) = (1 + 4 + 9) = (14)
Number of non-zero rows = 1
∴ P (AA^T) = 1

Question 2.
The rank of m × n matrix whose elements are unity is
(a) 0
(b) 1
(c) m
(d) n
Solution:
(b) 1

Question 3.
If T = is a transition probability matrix, then at equilibriuium A is equal to
(a) \(\frac { 1 }{ 4 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) \(\frac { 1 }{ 6 }\)
(d) \(\frac { 1 }{ 8 }\)
Solution:
(a) \(\frac { 1 }{ 4 }\)
Hint:

At equilibrium (A B) \(\left(\begin{array}{ll}
0.4 & 0.6 \\
0.2 & 0.8
\end{array}\right)\)
0.4 A + 0.2 B = A
0.4 A + 0.2(1 – A) = A
0.4 A + 0.2 – 0.2 A = A
0.2 A + 0.2 = A
0.2 = A – 0.2 A
0.2 = 0.8 A
A = \(\frac { 0.2 }{ 0.8 }\) = \(\frac { 1 }{ 4 }\)

Question 4.
If A = \(\left(\begin{array}{ll}
2 & 0 \\
0 & 8
\end{array}\right)\), then p(A) is
(a) 0
(b) 1
(c) 2
(d) n
Solution:
(c) 2
Hint:
A = \(\left(\begin{array}{ll}
2 & 0 \\
0 & 8
\end{array}\right)\)
No. of Non-zero rows = 2
∴ p(A) = 2

Question 5.
The rank of the matrix \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 9
\end{array}\right]\) is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3
Hint:

No. of Non-zero rows = 3
∴ p(A) = 3

Question 6.
The rank of the unit matrix of order n is
(a) n – 1
(b) n
(c) n + 1
(d) n²
Solution:
(b) n

Question 7.
If p(A) = r then which of the following is correct?
(a) all the minors of order r which does not vanish
(b) A has at least one minor of order r which does not vanish
(c) A has at least one (r + 1) order minor which vanishes
(d) all (r + 1) and higher order minors should not vanish
Solution:
(b) A has at least one minor of order r which does not vanish.

Question 8.
If A = \(\left(\begin{array}{l}
1 \\
2 \\
3
\end{array}\right)\) then the rank of AA^T is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(b) 1
Hint:

No. of non-zero rows = 1
p (AA^T) = 1

Question 9.
If the rank of the matrix \(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\) is 2. then λ is
(a) 1
(b) 2
(c) 3
(d) only real number
Solution:
(a) 1
Hint:
A = \(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\)
Since the Rank is 2, third order matrix vanishes
∴ |A| = 0
\(\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]\) = 0
λ(λ² – 0) + 1 (0 – 1) = 0
λ³ – 1 = 0 ⇒ λ³ = 1
∴ λ = 1

Question 10.

(a) 0
(b) 2
(c) 3
(d) 5
Solution:
(c) 3
Hint:
Number of non-zero rows = 3
∴ Rank = 3

Question 11.
If is a transition probability matrix, then the value of x is
(a) 0.2
(b) 0.3
(c) 0.4
(d) 0.7
Solution:
(c) 0.4
Hint:

is a transition probability matrix then 0.6 + x = 1
x = 1 – 0.6
x = 0.4

Question 12.
Which of the following is not an elementary transformation?
(a) R_i ↔ Rj
(b) R_i → 2R_i + 2Cj
(c) R_i → 2R_i + 4Rj
(d) C_i → C_i + 5Cj
Solution:
(b) R_i → 2R_i + 2Cj

Question 13.
If p(A) = p(A,B)= then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(d) Consistent

Question 14.
If p(A) = p(A,B)= the number of unknowns, then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(b) Consistent and has a unique solution

Question 15.
If p(A) ≠ p(A, B) =, then the system is
(a) Consistent and has infinitely many solutions
(b) Consistent and has a unique solution
(c) inconsistent
(d) consistent
Solution:
(c) inconsistent

Question 16.
In a transition probability matrix, all the entries are greater than or equal to
(a) 2
(b) 1
(c) 0
(d) 3
Solution:
(c) 0

Question 17.
If the number of variables in a non-homogeneous system AX = B is n, then the system possesses a unique solution only when
(a) p(A) = p(A, B) > n
(b) p(A) = p(A, B) = n
(c) p(A) = p(A, B) < n
(d) none of these
Solution:
(b) p(A) = p(A, B) = n

Question 18.
The system of equations 4x + 6y = 5, 6x + 9y = 7 has
(a) a unique solution
(b) no solution
(c) infinitely many solutions
(d) none of these
Solution:
(b) no solution
Hint:
The matrix form of the equations

p(A) ≠ p(A, B)
∴ The system is inconsistent.

Question 19.
For the system of equations x + 2y + 3z = 1, 2x + y – z = 3, 3x + 2y + k = 4
(a) there is only one solution
(b) there exists infinitely many solution
(c) there is no solution
(d) none of these
Solution:
(a) there is only one solution
Hint:
The matrix form of the equations

The last equivalent matrix in the echelon form
p(A) = p (A, B) = no. of unknowns
∴ The system is consistent.

Question 20.
If |A| ≠ 0, then A is
(a) non – singular matrix
(b) singular matrix
(c) zero matrix
(d) none of these
Solution:
(a) non-singular matrix

Question 21.
The system of linear equations x = y + z = 2, 2x + y – z = 3, 3x + 2y + k = 4 has unique solution, if k is not equal to
(a) 1
(b) 0
(c) 3
(d) 7
Solution:
(b) 0
Hint:
The matrix form of the equations

Since the system has unique solution.
P(A) = p(A, B) ≠ n
∴ K ≠ 0

Question 22.
Cramer’s rule is applicable only to get an unique solution when
(a) Δ_z ≠ 0
(b) Δ_x ≠ 0
(c) Δ ≠ 0
(d) Δ_y ≠ 0
Solution:
(c) Δ ≠ 0

Question 23.

Then (x, y) is

Solution:
(d) (\(\frac { -Δ_1 }{ Δ_2 }\), \(\frac { -Δ_1 }{ Δ_3 }\))

Question 24.
|A_{n × n}| = 3 |adj A| = 243 = (3)5 then the value n is
(a) 4
(b) 5
(c) 6
(d) 7
Solution:
(c) 6

Question 25.
Rank of a null matrix is
(a) 0
(b) -1
(c) ∞
(d) 1
Solution:
(a) 0

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 14 Marketing and Marketing Mix Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions 14 Marketing and Marketing Mix

12th Commerce Guide Marketing and Marketing Mix Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question 1.
The initial stage of Marketing system is ……………
a) Monopoly system
b) Exchange to money
c) Barter system
d) Self producing
Answer:
c) Barter system

Question 2.
Who is supreme in the market?
a) Customer
b) Seller
c) Wholesaler
d) Retailer
Answer:
a) Customer

Question 3.
In the following variables which one is not the variable of the marketing mix?
a) Place the variable
b) Product variable
c) Program variable
d) Price variable
Answer:
c) Program variable

Question 4.
Marketing mix means a marketing program that is offered by a firm to its target ………………… to earn profits through the satisfaction of their wants.
a) Wholesaler
b) Retailer
c) Consumer
d) Seller
Answer:
c) Consumer

Question 5.
Which one is the example of an Intangible product?
a) Education
b) Mobiles
c) Garments
d) Vehicles
Answer:
a) Education

II. Very Short Answer Questions.

Question 1.
What is Marketing?
Answer:
Marketing is one of the business functions that all activities that take place in relation to markets for actualise potential exchanges for the purpose of satisfying human needs and wants.

Question 2.
Define marketing mix.
Answer:
“Marketing mix is a pack of four sets of variables namely product variable, price variable, promotion variable, and place variable”

Question 3.
What is meant by Grading?
Answer:
Grading means the classification of standardized products into certain well-defined classes.

Question 4.
Define product.
Answer:
“A product is anything that can be offered to a market for attention, acquisition, use to consumption that might satisfy a want or a need” – PHILIP KOTLER

III. Short Answer Questions.

Question 1.
What are the objectives of marketing?
Answer:
Baker and Anshen say, “The end of all the marketing activities is the satisfaction of human wants”. The following are the objectives of marketing:

1. Intelligent and capable application of modem marketing policies.
2. To develop the marketing field.
3. To develop guiding policies and their implementation for a good result.

Question 2.
What is need for market and explain the concept of marketing?
Answer:
Market is needed for the following:

• To the movement of goods from the manufacturer to the consumer.
• To increase the standard of living of the people.
• To increase the national income
• To increase employment opportunities.

Concept of marketing:
What I can sell?

• Make what you can sell, but do not try to sell what you can make Shall I first create products?
• No, first create a consumer, then create products.

Shall I love my products?
No, love your customers not the products.

Who is supreme in the market?
Customer is supreme or king.
Who will shape my decision?
Customers’ preferences shape your decisions.

Question 3.
What are the factors affecting the Price of the Product? COMMO
Answer:
Factors affecting Price of product/service:
(a) Internal Factors:

1. Marketing Objectives
2. Marketing Mix Strategy
3. Organizational considerations
4. Costs
5. Organization Objectives

(b) External Factors:

1. The market after demand
2. Competition
3. Customers
4. Suppliers
5. Legal factors
6. Regulatory factors..

Question 4.
What do you mean by marketing mix? Describe any two elements.
Answer:

• Marketing mix means a marketing programme that is offered by a firm to its target consumers to earn profits through the satisfaction of their wants.
• It is a mixture of four ingredients namely 4Ps.
  • Product
  • Price
  • Place
  • Promotion.
  • Product
• “Product” is the main element of marketing.
• Without a product, there can be no marketing.
• Price is the value of a product expressed in monetary terms.
• t is the amount charged for the product.

IV. Long Answer Questions.

Question 1.
Discuss the evolution of marketing.
Answer:
Barter system:
Before the invention of money, goods were exchanged for goods.

Production orientation:

•  Producers more cared about the mass production of goods for the purpose of profit.
•  They cared very little about the customers.

Sales orientation:

• The stage witnessed a major change in all the spheres of economic life.
• Sales become the dominant factor, neglect the consumer needs and satisfaction.

Marketing orientation:
Customer’s importance was realized, but only as a means of disposing of goods aggressive advertising is used to compete for the stiff competition.

Consumer orientation:
Under this stage only such products are brought forward to the market which are capable of satisfying the tastes, preferences and expectations of consumer satisfaction.

Management orientation:
It assumes a managerial role to co-ordinate the business activities with the objective of planning, promoting and services to the present and potential consumers.

Question 2.
Why marketing is important to society and individual firms? Explain.
Answer:
To the Society:

•  Marketing is a connecting link between the consumers and the producers
•  Helps in increasing the living standard of people.
•  Helps in increasing the nation’s income.
•  Helps to increase employment opportunities.
•  II creates a modern cultivator.
•  Reduction in the cost of marketing is a direct benefit to society.
•  It includes all activities in the creation of Form, place, time, and possession utilities.

To the individual firms:

• It generates revenue [profit] to the firm.
• It helps the development of business and creates employment opportunities for people.
• It provides information to the top management for taking an overall decision on production
• Marketing and innovation are two basic functions of all business, the world is dynamic.

Question 3.
Narrate the elements of the marketing mix.
Answer:

Marketing mix means a marketing programme that is offered by a firm for the satisfaction of human wants. There are four elements of the marketing mix:

1. Product: A Product is the main element of marketing. Without a product, there can be no marketing.
2. Price: Price is the value of a product expressed in monetary terms. It is the amount charged for the product.
3. Place (Physical Distribution): An excellent quality product, with a good price, will be waste, if it is not transferred from the production place to the consumption place.
4. Promotion: An excellent product with a competitive price cannot achieve a desired success and acceptance in the market, with special features are conveyed to the consumers.

12th Commerce Guide Marketing and Marketing Mix Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
…………………………………….. Creates place utility.
a) Transport
b) Warehouse
c) Bank
d) Insurance
Answer:
a) Transport

Question 2.
Warehouse creates …………………………… utility.
a) Form
b) Place
c) Time
d) Possession
Answer:
c) Time

Question 3.
Before the invention of money, goods were exchanged for goods known as …………..
a) Trade
b) Commerce
c) Business
d) Barter
Answer:
d) Barter

Question 4.
Pick the odd one out:
a) Grading
b) Branding
c) Packing
d) Warehousing
Answer:
d) Warehousing

Question 5.
Pick the odd one out:
a) Price
b) Promotion
c) Place
d) Power
Answer:
d) Power

Question 6.
Choose the correct statement
i. Transport creates place utility.
ii. Were house creates time utility.
iii. Storage creates from utility

a) (i) and (ii) correct
b) (ii) and (iii) correct
c) All are correct
d) (i) and (ii) correct (iii) incorrect
Answer:
d) (i) and (ii) correct (iii) incorrect

Question 7.
Which one is not the correct method?
a) Promotion – Advertisement
b) Branding – Name
c) Packaging – Wrapping
d) Grading – Value
Answer:
d) Grading – value

Question 8.
Which of the following elements is the related marketing mix?
a) Place
b) production
c) consumption
d) trade
Answer:
a) place

II. Match The Following

Question 1.
Match list-1 with list-2

Answer:
a) (i) 4, (ii) 3, (iii) 2, (iv) 1

Question 2.
Match list-1 with list-2

Answer:
a) (i) 2, (ii) 4, (iii) 3, (iv) 1

III. Assertion and Reason.

Question 1.
Assertion [A]: Movement of goods from the place of production to the place of consumption by means of transport
Reason [R]: It creates place utility,
a) (A) is false (R) is true
b)(A) is true (R) is true
c) (A) is false (R) is false
d) (A) is true (R) is false
Answer:
b) (A) is True (R) is True

Question 2.
Assertion [ A]: Product is the main element of marketing.
Reason [R]: There can be no market without a product.
a) Both (A) and (R) are correct
b) Both (A) and (R) are incorrect
c) (A) is correct
d) (R) is correct
Answer:
a) Both (A) and (B) is correct

IV. Very Short Answer Questions.

Question 1.
Define marketing.
Answer:
“Marketing is concerned with the people and the activities involved in the flow of goods and services from the producer to the consumer”. -AMERICAN MARKETING ASSOCIATION.

Question 2.
What are the functions of exchange? BAS
Answer:

•  Buying
•  Assembling
•  Selling

Question 3.
Give the traditional marketing mix. 4Ps
Answer:

• Price
• Product
• Place
• Promotion

Question 4.
Give the modern marketing mix. 4Ps
Answer:

• People
• Performance
• Programme
• Process

Question 5.
What are the functions of marketing? G.B. GILES.
Answer:

•  Market research
•  Planning
•  Product development
•  Promotion
•  Distribution
•  Service offer sales

Question 6.
What are the functions of marketing? CLARK AND CLARK.
Answer:

•  Functions of exchange
•  Functions of physical supply
•  Facilitating functions.

Question 7.
Define price.
Answer:
“Price is the amount of money charged for a product or service or the sum of the values that consumers exchange for the benefits of having or using the product or service”. – PHILIP KOTLER

Question 8.
Give any two internal factors affecting the price of product/service. COMMO
Answer:

•  Costs
• Organizational objectives
•  Marketing objectives
•  Marketing mix strategy
•  Organizational consideration

V. Long Answer Questions.

Question 1.
Explain the functions of physical supply.
Answer:
Transportation:

•  Transport means carrying goods and services from one place to another. [Goods- men.]
•  It creates place utility by moving goods from the places when they are available in plenty to places where they are needed.
•  Types of transport – land, water, and air.

Storage:

•  Generally, there is a time gap between production and consumption of goods.
•  It involves the holding and preservation of goods from the time of production to the time of consumption.

Warehousing:
It creates time utility by storing the goods throughout the year and releasing them as and when they are needed.

Question 2.
What are the facilitating functions of marketing?
Answer:

• Financing: Long term – medium term and short term financing.
• Risk bearing: Time – place – competition – human and political risks.
• Grading: It means the classification of the standardized product into certain well-defined classes.
• Branding: It means giving a name or symbol to a product to differentiate it from a competitive product
• Packing: It means wrapping and crating of goods before distribution.
• Pricing: Price must be determined only to offer to take all the relevant factors into consideration.

Question 3.
What is marketing?
Answer:

• Marketing is one of the business functions that all activities that take place in relation to markets for actualizing potential exchanges [from the producer to the consumer] for the purpose of satisfying human needs and wants.
• Selling is basically concerned with putting the goods into the hands of the buyers for a price.
• But marketing is much wider than selling.
• It is an evolution rather than revolutionary.
• Marketing is what a marketer does. It is not clear to understand the meaning.
• It is one of the oldest professions in the world.
• The traditional objective of marketing had been to make the goods available at places where they are needed.
• This idea was later on changed by shifting the emphasis from exchange to satisfaction of human wants.
• Marketing must first find out what customers want and then plan a product to satisfy the wants.

Question 4.
State to advantages of warehousing.
Answer:
Storage:

• There is a time gap between production [supply] and consumption [demand].
• During the time gap, the goods may be damaged or destroyed or perished.
• For that goods need storage or warehouse.
• It creates time utility.

Equalization of demand and supply:

• It equalizes the demand and supply of goods.
• Storing the goods- when there is no demand.
• Releasing the goods – when there is a demand.

Price stabilization:

• It ensures price stabilization.
• It controls the price fluctuations.
• The price of commodities more or less uniform throughout the year by storing the goods, if there is no demand and releasing the goods if they are demanded.

Risk bearing:
When the goods are in the warehouse, in case of damage due to flood fire, theft, etc., the warehouse keepers compensate the loss of the owner of the goods.

Finance:
By pledging warehouse receipt or warehouse warrant, a depositor can get loans from financial institutions.

Question 5.
How market information is helpful to the invention of a new product in the market?
Answer:

•  Strong knowledge of customers.
•  Marketing research – process for marketing opportunities- problem-solving collecting- analyzing marketing information.
• Control over fact – the responsibility of the marketing research director.
• Design research carefully and supervise its execution.
• Various combinations of people, technologies for managing marketing information.
• Broader management and strategy decisions.
• Research consumer performance to contribute to improving decision-making performance.
• It is helpful to accurate needs and decision-making by the marketer to be a regular source of information about marketing and trading brand awareness.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 13 Concept of Market and Marketer Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 13 Concept of Market and Marketer

12th Commerce Guide Concept of Market and Marketer Text Book Back Questions and Answer

I. Choose The Correct Answer.

Question 1.
One who promotes (or) Exchange of goods or services for money is called as ……………
a) Seller
b) Marketer
c) Customer
d) Manager
Answer:
b) Marketer

Question 2.
The marketer initially wants to know in the marketing is …………………….
a) Qualification of the customer
b) Quality of the product
c) Background of the customers
d) Needs of the customers
Answer:
d) Needs of the customers

Question 3.
The Spot market is classified on the basis of ………………..
a) Commodity
b) Transaction
c) Regulation
d) Time
Answer:
b) Transaction

Question 4.
Which one of the market deals in the purchase and sale of shares and debentures?
a) Stock Exchange Market
b) Manufactured Goods Market
c) Local Market
d) Family Market
Answer:
a) Stock Exchange Market

Question 5.
Stock Exchange Market is also called …………
a) Spot Market
b) Local Market
c) Security Market
d) National Market
Answer:
c) Security Market

II. Very Short Answer Questions.

Question 1.
What is the Market?
Answer:
The word market is derived from the Latin word ‘Marcatus’ which means trade, commerce, merchandise, a place where the business is transacted.

Question 2.
Define Marketer.
Answer:
“A person whose duties include the identification of the goods and services desired by a set of consumers, as well as the marketing of those goods and services on behalf of a company” – BUSINESS DICTIONARY

Question 3.
What is meant by Regulated Market?
Answer:
Very Short Period Market: Markets that deal in perishable goods like, fruits, milk, vegetables, etc., are called a very short period market.

Question 4.
What is meant by Spot Market?
Answer:
In such markets, goods are exchanged and the physical delivery of goods takes place immediately.

Question 5.
What is meant by Commodity Market?
Answer:
A commodity market is a place where produced goods or consumer goods are bought and sold.

III. Short Answer Questions.

Question 1.
What can be marketed in the Market? [S.GOPI PEEPI]
Answer:

• Services
• Goods
• Organization
• Persons
• Ideas
• Places
• Events
• Experiences
• Properties
• Information

Question 2.
Mention any three Roles of the Marketer.
Answer:
Instigator:
As an Instigator, He keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever-changing market.

Innovator:

• Marketer seeks to distinguish his product or services by adding additional features or functionalities to the existing products.
• Modifying the pricing structure, introducing new delivery patterns, creating new business models, introducing changes in the production process, and so on.
• Implementer: Marketer plays the role of implementer when he or she actually converts marketing opportunities into marketable products with the help of several functional teams put in place in the organization.

Question 3.
The marketer is an innovator? Do you agree?
Answer:
Marketer seeks to distinguish his products/services by adding additional features or functionalities to the existing product, modifying the pricing structure, introducing new delivery patterns, creating new business models, introducing change in production process and so on.

Question 4.
Explain the types of markets on the basis of time.
Answer:
Very Short Period Market:
Markets which deal in perishable goods like milk, Fish, Fruits, Vegetables, etc. are called as “Very Short Period Market”.

Short Period Market:
In certain goods, supply is adjusted to meet the demand. The demand is greater than supply. (Rice, wheat, etc) such markets are known as “Short Period Marker.

Long Period Market:
This type of market deals in durable goods. [TV – Fridge – computer etc] where the goods and services are dealt for longer period usages.

Question 5.
List down the functions of Marketer? [GM] [5-P] [BTSS]
Answer:

1. Very Short Period Market: Markets which deal in perishable goods like, fruits, milk, vegetables, etc., are called a very short period market. There is no change in the supply of goods.
2. Short Period Market: In certain goods, supply is adjusted to meet the demand. The demand is greater than supply. Such markets are known as Short Period Market.
3. Long Period Market: This type of market deals in durable goods, where the goods and services are dealt with for a longer period of usages.

IV.Long Answer Questions.

Question 1.
How the Market can be classified? GTTV RICE
Answer:

Question 2.
How the market can be classified on the basis of Economics?
Answer:
The Market can be classified on the basis of Economics as follows:
(a) Perfect Market: A market is said to be a perfect market if it satisfies the following conditions:

1. A large number of buyers and sellers are there.
2. Prices should be uniform throughout the market.
3. Buyers and sellers have a perfect knowledge of the market.
4. Goods can be moved from one place to another without restrictions.

(b) Imperfect Market: A market is said to be imperfect when

1. Products are similar but not identical.
2. Prices are not uniform.
3. There is a lack of communication.

12th Commerce Guide Concept of Market and Marketer Additional Important Questions and Answers

I. Choose The Correct Answer.

Question 1.
One who promotes (or) Exchange of goods or services for money is called as _____
(a) Seller
(b) Marketer
(c) Customer
(d) Manager
Answer:
(b) Marketer

Question 2.
Mercatus Mean ……….
a) A place where the business is transacted
b) A place where goods are produced
c) A place where manufacturers assembled
d) All of these
Answer:
a) A place where the business is transacted

Question 3.
The Need for the market is ……………………..
a) To Exchange
b) To adjust the price mechanism
c) To improve the quality
d) All the above
Answer:
d) All the above

Question 4.
Which one of the market deals in the purchase and sale of shares and debentures?
(a) Stock Exchange Market
(b) Manufactured Goods Market
(c) Local Market
(d) Family Market
Answer:
(a) Stock Exchange Market

Question 5.
In …………………. market physical delivery of goods takes place immediately.
a) Future market
b) Spot market
c) Capital market
d) Commodity market
Answer:
b) Spot market

Question 6.
Short term securities are exchanged in ……………… market.
a) Stock
b) Finance
c) Money
d) All of the above
Answer:
c) Money

Question 7.
Stock Exchange Market is also called _____
(a) Spot Market
(b) Local Market
(c) Security Market
(d) National Market
Answer:
(d) National Market

Question 8.
Which one is not correctly matched?
a) Very short period market – Vegetables
b) Future market – Delivery
c) Regulated market – No statutory measures
d) Bullion market – Gold
Answer:
c) Regulated market – No statutory measures

Question 9.
Which of the following statement correct?
i) In imperfect market products are similar but not identical.
ii) Prices are not uniform.
iii) There is a lack of communication.

a) (i) is correct
b) (ii) and (iii) are correct
c) (i) and (ii) are correct
d) All (i), (ii) and (iii) are correct
Answer:
d) All (i), (ii) and (ii) are correct

II. Match The Following.

Question 1.
Match List-I with List-II

a) (i) – 2 (ii) – 1 (iii) – 4 (iv) – 3
b) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
c) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
d) (i) – 1 (ii) – 4 (iii) – 3 (iv) – 2
Answer:
a) (i) – 2 (ii) – 1 (iii) – 4 (iv) – 3

Question 2.
Match List-I with List-II

a) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 1
b) (i) – 3 (ii) – 2 (iii) – 1 (iv) – 4
c) (i) – 2 (ii) – 3 (iii) – 4 (iv) – 1
d) (i) – 1 (ii) – 4 (iii) – 3 (iv) – 2
Answer:
a) (i) – 4 (ii) – 3 (iii) – 2 (iv) – 4

III. Assertion and Reason.

Question 1.
Assertion (A): A market is said to be a perfect market.
Reason (R): There is a Large number of buyers and sellers.
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are True
d) Both (A) and (R) are False
Answer:
c) Both (A) and (R) are True

IV. Short Answer Questions.

Question 1.
Mention any three Roles of the Marketer.
Answer:

1. Instigator: As an instigator, the marketer keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever-changing market.
2. Integrator: The marketer plays the role of the integrator in the sense that he collects feedback or vital inputs from channel members and consumers.
3. Implementer: Marketer plays the role of implementer when he/she actually converts marketing opportunities into marketable products.

Question 2.
Classify the market on the basis of Importance.
Answer:
Primary market:
The primary producers of firms sell their output or products through this type of market to wholesales or consumers.

Secondary market:

• In this market, semi-finished goods are marketed.
• The dealings commonly between wholesales and retailers.

Terminal market:
It is a central place that serves as an assembly and trading place for commodities in a
metropolitan area.

Question 3.
List down the functions of Marketer.
Answer:

1. Gathering and Analysing market information
2. Market planning
3. Product Designing and development
4. Standardization and Grading
5. Packaging and Labelling
6. Branding
7. Customer Support Services
8. Pricing of Products
9. Promotion and Selling
10. Physical Distribution
11. Transportation
12. Storage and Warehousing

Question 4.
Define Market.
Answer:
“Market includes both place and region in which buyers and sellers are in free competition with one another” – PYLE.

Question 5.
Explain the Need for marketing.
Answer:

•  To Exchange [Barter] Goods and Services
•  To Adjust Demand and Supply by the price mechanism.
•  To improve the quality of life of the Society.

Question 6.
Mention any four differences between Wholesale Market and Retail Market? [LOQS]
Answer:

Question 7.
Why Customer support is needed to Market?
Answer:

•  To exchange (Barter) goods and services.
•  To adjust demand and supply by the price mechanism.
•  To improve the quality of life of the society, [standard of living]
•  To introduce new modes of life.
•  To develop by enhancing the market segment.

V. Long Answer Questions.

Question 1.
How the market can be classified?
Answer:
On the basis of different approaches markets can be classified as follows:
I. On the basis of Geographical Area:

• Family Market
• Local Market
• National Market
• International Market or World Market

II. On the Basis of Commodities / Goods:
(a) Commodity Market

• Produce Exchange Market
• Manufactured Goods Market
• Bullion Market

(b) Capital Market:

• Money Market
• Foreign Exchange Market
• The Stock Market

III. On the Basis of Economics:

• Perfect Market
• Imperfect Market

IV. On the basis of transaction:

• Spot Market
• Future market

V. On the Basis of Regulation:

• Regulated Market
• Unregulated Market

VI. On the Basis of Time:

• Very Short Period Market
• Short Period Market
• Long Period Market

VII. On the Basis of Volume of Business:

• Wholesale Market
• Retail Market

VIII. On the Basis of Importance:

• Primary Market
• Secondary Market
• Terminal Market

Question 2.
How the market can be classified on the basis of commodities [Goods]?
Answer:
A) Commodity Market:
A commodity market is a place where produced goods or consumer goods are bought and sold.

1. Produce Exchange Market: It is an organised market where commodities or Agricultural produce are Bought and sold on a wholesale basis.
2. Manufactured Goods Market: This market deals with [Leather goods, machinery, etc] manufactured goods.
3. Bullion Market: This market deals with the purchase and sale of Gold and Silver.

B) Capital Markets:

1. Money Market: It is a type of market where short term securities are exchanged. [Banks, – Finance to industries] .
2. Foreign Exchange Market: This type of market, help Exporters and Importers, in converting their currencies into Foreign currencies and vice versa.
3. Stock Market: This is a market where the purchase and sales of shares and Debentures, Bonds, etc. of companies are dealt with.

Question 3.
What is your contribution to promote the market in modern society?
Answer:
A market is a place where buyers and sellers gather for purchase and sale. The market may be of Local market, national market, and international or global market. To develop and promote the market the following are needed:

1. Eligible and satisfied and customer is needed.
2. Quality and durable goods are to be marketed.
3. Recent trends like E-marketing, online marketing are to be encouraged.
4. After-sales service is to be provided to durable goods.
5. Customers are to be financed for buying costly articles.
6. New Innovations and marketing research are to be introduced to develop the market.

Question 4.
Apart from goods and services, explain any five items that can be marketed,
Answer:
i) Experiences:
The unique and varied experiences pertaining to a place or a park or an event can be marketed under this concept. For eg. Amusement Park, Theme Park, Mountaineering, etc.

ii) Events:
The event marketing aims at promoting and marketing special events, shows, exhibitions, fairs, performances, sports events like the World Cup, Olympics, T20, etc.

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 10 Surface Chemistry Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 10 Surface Chemistry

12th Chemistry Guide Surface Chemistry Text Book Questions and Answers

Part – I Text Book Evaluation

I. Choose the correct answer

Question 1.
For freudlich isotherm a graph of log \(\frac{x}{m}\) is plotted against log P. The slope of the line and its y – axis intercept respectively corresponds to
(a) \(1 / n\), k
(b) log \(1 / n\), k
(c) \(1 / n\), log k
(d) log \(1 / n\), log k
Answer:
(c) \(1 / n\), log k
\(\frac{x}{m}\) = \(\mathrm{k} \cdot \mathrm{p}^{1 / \mathrm{n}}\)
log\((\frac{x}{m})\) = log k + \(\frac { 1 }{ n }\)log p
y = c + mx
m = \(\frac { 1 }{ n }\) and c = log k

Question 2.
Which of the following is incorrect for physisorption?
(a) reversible
(b) increases with increase in temperature
(c) low heat of adsorption
(d) increases with increase in surface area
Answer:
(b) increases with increase in temperature

Question 3.
Which one of the following characteristics are associated with adsorption?
(a) ∆G and ∆H are negative but ∆S is positive
(b) ∆G and ∆S are negative but ∆H is positive
(c) ∆G is negative but ∆H and ∆S are positive
(d) ∆G. AH and ∆S all are negative.
Answer:
(d) ∆G, ∆H and ∆S all are negative.
Adsorption leads to decrease in randomness (entropy).i.e. ∆S < 0 for the adsorption to occur, ∆G should be – ve. We know that ∆G = ∆H – T∆S if ∆S is – ve, T∆S is + ve. It means that ∆G will become negative only when ∆H is – ve and ∆H > T∆S

Question 4.
Fog is colloidal solution of ……………..
(a) solid in gas
(b) gas in gas
(c) liquid in gas
(d) gas in liquid
Answer:
(c) liquid in gas
dispersion medium-gas, dispersed phase-liquid

Question 5.
Assertion: Coagulation power of Al³⁺ is more than Na.
Reason: greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechuize nile)

Question 6.
Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?
(a) It is not true, ferric chloride is a poison.
(b) It is true, Fe³⁺ ions coagulate blood which is a negatively charged sol
(c) It is not true; ferric chloride is ionic and gets into the blood stream.
(d) It is true, coagulation takes place because of formation of negatively charged sol with Cl^–.
Answer:
(b) It is true, Fe³⁺ ions coagulate blood which is a negatively charged sol

Question 7.
Hair cream is …………..
(a) gel
(b) emulsion
(c) solid sol
(d) sol.
Answer:
(b) emulsion
Emulsion dispersed phase, Dispersion medium -liquid

Question 8.
Which one of the following is correctly matched?
(a) Emulsion – Smoke
(b) Gel – butter
(c) foam – Mist
(d) whipped cream – sol
Answer:
(b) Gel – butter

Question 9.
The most effective electrolyte for the coagulation of As₂S₃ Soils
(a) NaCl
(b) Ba(NO₃)₂
(c) K₃[Fe(CN)₆]
(d) AI₂(SO₄)₃
Answer:
(d) AI₂(SO₄)₃

Question 10.
Which one of the is  not a surfactant?
(a) CH₃ – (CH₂)₁₅ – N – (CH₃)₂CH₂Br
(b) CH₃ – (CH₂)₁₅ – NH₂
(c) CH₃ – (CH₂)₁₆ – CH₂OSO₂ – Na⁺
(d) OHC – (CH₂)₁₄ – CH₂ – COO^–Na⁺
Answer:
(b) CH₃ – (CH₂)₁₅ – NH₂

Question 11.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………….
(a) Cataphoresis
(b) Electrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect-scattering of light

Question 12.
In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K₂SO₄
(i). Na₃PO₄
(ii). K₄[Fe(CN)₆]
(iii). and NaCI
(iv). Their coagulating power should be …………..
(a) II > I >IV > III
(b) III > II > I > IV
(c) I > II > III > IV
(d) none of these
Answer:
(b) III > II > I > IV

Question 13.
Collodion is a 4% solution of which one of the following compounds in alcohol – ether mixture?
(a) Nitroglycerine
(b) Cellulose acetate
(c) Glycoldinitrate
(d) Nitrocellulose
Answer:
(a) Nitrocellulose
pyroxylin (nitro cellulose)

Question 14.
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber’s process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(a) Hydrolysis of sucrose in presence of all HCI
Answer:
(a) Hydrolysis of sucrose in presence of all HCl
Both reactant and catalyst are in same phase. i.e. (1)

Question 15.
Match the following.

Answer:
(a) (iv) (i) (ii) (iii)

Question 16.
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS₂S₃ are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO₄) = 0.22
The correct order of their coagulating power is ……….
(a) III > II > I
(b) I > II > III
(c) I >III > II
(d) II > III > I
Answer:
(a) III > II > I

Question 17.
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then ……………
(a) ∆H increases
(b) ∆S increases
(c) ∆G increases
(d) ∆S decreases
Answer:
(a) ∆S decreases – ∆S is -ve

Question 18.
If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?
(a) x/m = f(P) at constant T
(b) x/m = f(T) at constant P
(c) P = f(T) at constant x/m
(d) x/m = PT
Answer:
(d) x/m = mPT

Question 19.
On which of the following properties does the coagulating power of an ion depend?
(a) Both magnitude and sign of the charge on the ion.
(b) Size of the ion alone
(c) the magnitude of the charge on the ion alone
(d) the sign of charge on the ion alone.
Answer:
(a) Both magnitude and sign of the charge on the ion.

Question 20.
Match the following.

II. Short Answer Questions

Question 1.
Give two important characteristics of physisorption.
Answer:
Important characteristics of physisorption:

1. It is reversible
2. It has low heat of adsorption
3. It has weak van der Waals forces of attraction with the adsorbent.
4. It increases with an increase in pressure.
5. It forms multi molecular layer.

Question 2.
Differentiate physisorption and chemisorption.
Answer:
Chemical adsorption or Chemisorption or Activated adsorption

1. It is very slow
2. It is very specific depends on nature of adsorbent and adsorbate.
3. chemical adsorption is fast with increase pressure, it can not alter the amount.
4. When temperature is raised chemisorption first increases and then decreases.
5. Chemisorption involves transfer of electrons between the adsorbent and adsorbate, Heat of adsorption is high i.e., from 40 – 400kJ/mole.
6. Monolayer of the adsorbate is formed.
7. Adsorption occurs at fixed sites called active centres. It depends on surface area.
8. Chemisorption involves the formation of activated complex with appreciable activation energy.
9. Physical adsorption or van der Waals adsorption or Physisorptlon
10. It is irreversible.

Physical adsorption or van der Waals adsorption or physisorption.

1. It is instantaneous
2. It is non-specific
3. In Physisorption. when pressure increases the amount of adsorption increases.
4. Physisorption decreases with increase in temperature.
5. No transfer of electrons
6. Heat of adsorption is low in the order of 40kJ/mole.
7. Multilayer of the adsorbate is formed on the adsorbent.
8. It occurs on all sides.
9. Activation energy is insignificant.
10. It is reversible.

Question 3.
In case of chemisorption, why adsorption first increases and then decrease with temperature?
Answer:
1. Chemisorption involves high activation energy so it is also referred to as activated adsorption.

2. It is found in chemisorption that it first increases and then decreases with an increase in temperature. When adsorption is plotted, the graph first increases and then decreases with temperature.

3. The initial increase illustrates the requirement of activation of the surface for adsorption is due to fact that the formation of activated complex requires certain energy. But later it decreases at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)

Question 4.
Which will be adsorbed more readily on the surface of charcoal and why; NH₃ or CO₂?
Answer:
1. The gases having low critical temperature are adsorbed slowly, while gases with high critical temperature are adborbed readily.

2. Among CO₂, and NH₃, NH₃ will be more readily adsorbed on the surface of the charcoal. This is because the critical temperature of ammonia gas is quite high than the CO₂. Hence, it easily combines with the materials than the CO₂ whether it is solid, liquid or any gases.

Question 5.
Heat of adsorption is greater for chemisorptions than physisorption. Why?
Answer:
Chemisorption has higher heat of adsorption. because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chenil sorptions than physisorption. Chemisorption, heat of adsorption range 40 – 400kJ/mole.

Question 6.
In a coagulation experiment 10 mL of a colloid (X) is mixed with distilled water and 0.1M solution of an electrolyte AB so that the volume is 20 mL. It was found that all solutions containing more than 6.6 mL of AB coagulate with in 5 minutes. What is the flocculation values of AB for sol (X)?
Answer:
A minimum of 6.6mL of AB is required to coagulate the sol. The moles of AB in the sol is
\(\frac{6.6 \times 0.01}{20}\) = 0.033 moles
This means that a minimum of 0.033 moles or 0.0033 x 1000 = 3.3 milli moles are required for coagulating one litre of sol. Flocculation value of AB for X = 3.3

Question 7.
Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.
Answer:
1. Ions either positive or negative of peptizing agent (electrolyte) are adsorbed on the particles of precipitate. They repel and hit each other and break the particles of the precipitate into colloidal size.

2. For example, when we add a small volume of very dilute hydrochloric acid solution peptising agent to a fresh precipitate of a silver chloride, it leads to formation of silver chloride colloidal solution,

Question 8.
What happens when a colloidal sol of Fe(OH)₃ and As₂S₃ are mixed?
Answer:
On mixing Fe(OH)₃ positive sol and As₂S₃ negative sol, mutual coagulation occurs which causes precipitation. When this sol got mixed with each other, due to Fe³⁺ and S^2- ions neutralisation of charges will happen, and precipitate will be formed.
Fe(OH)₃ + As₂S₃ → Fe₂S₃ + As(OH)₃

Question 9.
What is the difference between sol and gel?
Answer:
Sol

1. The liquid state of a colloidal solution is called a sol.
2. The sol does not have a definite structure.
3. The dispersion medium of the sol may be water.
4. The sol can be converted to gel by cooling The sol can be easily dehydrated.
5. The viscosity of the sol is very low.
6. Sol is categorized into lyophobic and lyophilic sols.
7. Example: Blood

Gel

1. The solid or semi-solid state of a colloidal solution is called gel.
2. The gel possesses a honeycomb-like structure.
3. The dispersion medium of gel will be hydrated colloid particles.
4. The gel can be converted to sol by heating.
5. The gel cannot be dehydrated.
6. The viscosity of the gel is very high.
7. There is no such classification of gel.
8. Example: Fruit jelly, cooked gelatin jelly.

Question 10.
Why are lyophillic colloidal sols are more stable than lyophoblic colloidal sol?
Answer:
1. A lyophilic colloidal sols are stable due to the charge and the hydration of sol particles.

2. Lyophilic sols are more stable than lyophobilc sols because they are highly hydrated in the solution. And since more is the hydration more will be its stability.

3. Lyophilic sols are stabilized by electrostatic charge and hydration whereas lyophobic sols are only stabilized by charge, so they easily gets coagulated and requires a stabilising agent. Hence, lyophilic sols are more stable than lyophobilc sols.

Question 11.
Addition of Alum purifies water. Why?
Answer:
Purification of drinking water is activated by coagulation of suspended impurities in water using alums containing Al³⁺. That is why we are adding to purify water.

Question 12.
What are the factors which influence the adsorption of a gas on a solid?
Answer:
Factors that influence the adsorption of a gas on a solid is as follows:
1. Nature of the gas:
Easily liquefiable gases such as NH₃, HCl etc are adsorbed to a great extent in comparison to gases such as H₂, O₂ etc. This is because van der Waal’s forces are stronger is easily liquifiable gases.

2. Surface area of the solid:
The greater the surface area of the adsorbent, the greater is the adsorption of gas on the solid surface.

3. Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

4. Effect of temperature:
Adsorption is an exothermic process. Thus in accordance with Le – Chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 13.
What are enzymes? Write a brief note on the mechanism of enzyme catalysis.
Answer:
Enzymes are complex protein molecules with three-dimensional structures. They catalyse the chemical reaction in a living organism. They are often present in colloidal state and extremely specific in catalytic action.

Each enzyme produced in a particular living cell can catalyse a particular reaction in the cell. Mechanism of enzyme catalysis: Mechanism of enzyme catalysed reaction is known as lock and key mechanism.
1. Enzymes arc highly specific in their action.

2. This specificity is due to the pressure of active sites. The shape of active site of any given enzyme is like cavity such that only a specific substrate can fit into it.

In the same way a key fits into lock. The specific binding needs to the formation of an enzyme-substrate complex which accounts for high specificity of enzyme catalysed reactions.

3. Once the proper orientation is attained the substrate molecules reacts to form the product in two steps.

4. Since product molecules do not have any affinity for the enzyme they leave the enzyme surface making room for fresh substrate.

step 1: Formation of the enzyme-substrate complex

Step 2: Dissociation of the enzyme-substrate complex to form product

The rate of the formation of the product depends upon the concentration of ES.

Question 14.
What do you mean by activity and selectivity of catalyst?
Answer:
1. Activity of Catalyst:
The activity of a catalyst is its ability to increase the rate of a particular reaction, Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants in the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

2. Selectivity of the catalyst:
The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H₂ and CO.

Question 15.
Describe some features of catalysis b Zeoliles.
Answer:
1. Zeolites are microporous, crystalline, hydrated aluminosilicates, made of silicon and aluminium tetrahedra.

2. There are about 50 natural zeolites and 1 50 synthetic zeolites. As silicon is tetravalent and aluminium is trivalent, the zeolite matrix carries extra negative charge. To balance the negative charge, there are extra framework cations for example, H⁺ or Na⁺ ions.

3. Zeolites earring protons are used as solid acids, catalysis and they are extensively used in the petrochemical industry for cracking heavy hydrocarbon fractions into gasoline, diesel,etc.

4. Zeolites earring Na ions are used as basic catalysis.

5. One of the most important applications of zeolites is their shape selectivity. In zeolites, the active sites namely protons are lying inside their pores. So, reactions occur only inside the pores of zeolites.

Question 16.
Give three uses of emulsions.
Answer:

1. The cleansing action of soap is due to emulsions.
2. It is used in the preparation of vanishing cream.
3. It is used in the preparation of cold liver oil.

Question 17.
Why does the bleeding stop by rubbing moist alum?
Answer:
Blood is a colloidal sol. When we nib the injured part with moist alum then coagulation of blood takes place. Hence main reason is coagulation, which stops the bleeding. Therefore bleeding stop by rubbing moist alum.

Question 18.
Why is desorption important for a substance to act as good catalyst?
Answer:
Desorption is important for a substance to act as a good catalyst, so that after the reaction, the products found on the surface separate out (desorbed) to create free surface again for other reactant molecules to approach the surface and react. If desorption does not occur then other reactants are left with no space on the surface of the catalyst for adsorption and the reaction will stop.

Question 19.
Comment on the statement: Colloid is not a substance but it is a state of the substance.
Answer:
The statement is true. Because the same substance may exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example. NaCl in water behaves as a crystalloid while in benzene, it behaves as a colloid. Similarly, dilute soap solution behaves

like a crystalloid while concentrated solutions behave as a colloid. It is the size of the particles which matters. That is the state in which the substance exists. If the size of the particles lies in the range 1 nm to 1oo nm, it is in the colloidal state.

Question 20.
Explain any one method for coagulation
Answer:
The flocculation and setting down of the soil particles is called coagulation. Various method of coagulation are given below:

1. Addition of electrolytes
2. Electrophoresis
3. Mining oppositely charged sols
4. Boiling.

Addition of electrolytes
A negative ion causes the precipitation of positively charged sol and vice versa. When the valency of ion is high, the precipitation power is increased. For example, the precipitation power of some cations and anions varies in the following order
Al³⁺ > Ba²⁺ > Na⁺, Similarly [Fe(CN)₆]^-3 > SO₄^-2 > Cl^–
The precipitation power of electrolyte is determined by finding the minimum concentration (millimoles / lit) required to cause precipitation of a sol in 2hours. This value is called flocculation value. The smaller the flocculation value greater will be precipitation.

Question 21.
Write a note on electroosmosis.
Answer:
Electro osmosis:
A sol is electrically neutral. Hence the medium carries an equal but opposite charge to that of dispersed particles. When sol particles are prevented from moving, under the influence of electric field the medium moves in a direction opposite to that of the sol particles. This movement of dispersion medium under the influence of electric potential is called electro-osmosis.

Question 22.
Write a note on catalytic poison
Answer:
Catalytic poison:
Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons. For example, In the reaction,
2SO₂ + O₂ → 2SO₃ with a Pt catalyst, the poison is AS₂O₃.
i.e., AS₂O₃ destroys the activity of pt. AS₂O₃ blocks the activity of the catalyst. So, the activity is lost.

Question 23.
Explain intermediate compound formation theory of catalysis with an example.
Answer:
The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation. in homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the catalyst is regenerated.

Consider the reactions:
A + B → AB ……………(1)
A + C → AC (intermediate) ………….(2)
C is the catalyst
AC + B → AB + C …………(3)
Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.
Example:
The mechanIsm of Fridel crafts reaction is given below

The action of catalyst is explained as follows .

This theory describes,

1. The specificity of a catalyst.
2. The increase in the rate of the reaction with increase in the concentration of a catalyst.

Limitations

1. The intermediate compound theory fails to explain the action of catalytic poison and activators (promoters).
2. This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 24.
What is the difference between homogenous and hetrogenous catalysis?
Answer:
Hornogenous Catalysis:

1. In a catalysed reaction the reactants, products and catalyst are present in the same phase.
2. For example.
   
   Hence NO act as catalyst.
3. Homogeneous catalysis explained by intermediate compound formation theory.

Heterogeneous Catalysis:

1. In a reaction, the catalyst is present in a different phase. i.e., catalyst is not present in the same phase as that of reactants and products.
2. For example.
   
   Hence Pt(s) act as catalyst.
3. Hetenogeneous catalysis explained by adsorption theory.

Question 25.
Describe adsorption theory of catalysis.
Answer:
Langmuir explained the action of catalyst in heterogeneous catalysed reactions based on adsorption.
This theory explains heterogeneous catalysis.
The reactant molecules are adsorbed on the catalytic surface, so heterogeneous catalysis is also called as contact catalysis.
When the reactants are adsorbed on the catalytic surface, an activated complex is formed.
This activated complex decomposes to give the product.
Steps involved in heterogeneous catalysis.

1. Reactant molecules diffuse from bulk to the catalyst surface.
2. Reactant molecules are adsorbed on the surface of the catalyst.
3. Adsorbed reactant molecules are activated to form activated complex which is decomposed to form the products.
4. Product molecules are desorbed.
5. Product molecules diffuse away from the surface of the catalyst.

12th Chemistry Guide Surface Chemistry Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
Adsorption is a
a) Bulk phenomenon
b) Surface phenomenon
c) Both (a) and (b)
d) None of the above
Answer:
b) Surface phenomenon

Question 2.
Absorption is a
a) Bulk phenomenon
b) Surface phenomenon
c) Both (a) and (b)
d) None of the above
Answer:
a) Bulk phenomenon

Question 3.
Which among the following is an adsorbent?
a) N₂
b) SO₂
c) Ni
d) NH₃
Answer:
c) Ni

Question 4.
In adsorption, if the concentration of a substance in the interface is high, then it is called?
a) Desorption
b) Positive adsorption
c) Negative adsorption
d) Absorption
Answer:
b) Positive adsorption

Question 5.
Adsorption process is
a) Spontaneous
b) Non-spontaneous
c) Slow
d) Bulk phenomenon
Answer:
a) Spontaneous

Question 6.
Adsorption is always accompanied by
a) Increase in entropy
b) Increase in free energy
c) Decrease in free energy
d) No change in entropy
Answer:
c) Decrease in free energy

Question 7.
The force of attraction exist between adsorbent and adsorbate in physical adsorption is
a) Vanderwaal’s force
b) Dipole-dipole interaction
c) Dispersion forces
d) All the above
Answer:
d) All the above

Question 8.
Total amount of the gas adsorbed increases as the ……………. of the adsorbent increases.
a) Volume
b) Density
c) Surface area
d) Surface tension
Answer:
c) Surface area

Question 9.
The critical temperature of the gas which is readily adsorbed is
a) lower
b) higher
c) zero
d) none of the above
Answer:
b) higher

Question 10.
Which among the following gas is adsorbed slowly?
a) SO₂
b) NH₃
c) N₂
d) CO₂
Answer:
c) N₂
Reason: Critical temperature of N₂ is low hence adsorbed slowly.

Question 11.
The process of adsorption is
a) Exothermic
b) Endothermic
c) Both (a) and (b)
d) None of the above
Answer:
a) Exothermic

Question 12.
Multi molecular layers are formed in
a) Absorption
b) Physisorption
c) Chemisorption
d) None of the above
Answer:
b) Physisorption

Question 13.
The rate constant of a reaction at temperature 200K is 10 times less than the rate constant at 400K. What is the activation energy of the reaction? (R=gas constant)
a) 1842.4R
b) 921.2 R
c) 460.6 R
d) 230.3 R
Answer:
b) 921.2 R

Question 14.
In adsorption isobar the amount of adsorption is plotted against
a) Pressure
b) Temperature
c) Volume
d) Mass
Answer:
b) Temperature

Question 15.
In physical adsorption isobar x /m, …………………………. with increase in temperature.
a) Increases
b)Decreases
c) First increases then decreases
d) Does not change
Answer:
b) Decreases

Question 16.
In chemisorption isobar x /m …………….. with increase in temperature.
a) Increases
b)Decreases
c) First increases then decreases
d) Does not change
Answer:
c) First increases then decreases

Question 17.
Which of the following is not an equation for Freundlich isotherm?

Answer:
d

Question 18.
Sugar prepared from molasses in decolourised by adding.
a) Silica gel
b) Permutit
c) Animal Charcoal
d) Activated charcoal
Answer:
c) Animal Charcoal

Question 19.
Which of the following is not an application of adsorption?
a) Heterogeneous catalysis
b) Gas masks
c) Froth floatation process
d) Softening of water by boiling
Answer:
d) Softening of water by boiling

Question 20.
The change of W/O emulsion into O/W emulsion is called …………..
a) Coagulation
b) Emulsification
c) Decomposition
d) Invension of phase
Answer:
d) Invension of phase

Question 21.
2SO₂ + O₂ + [NO] → 2SO₃ + [NO] is an example for
a) Positive catalysis
b) Negative catalysis
c) Homogeneous catalysis
d) Both (a) and (c)
Answer:
d) Both (a) and (c)

Question 22.
Which is not a heterogeneous catalysis?
a) Contact process
b) Haber’s process
c) Ester hydrolysis
d) Ostwald’s process
Answer:
c) Ester hydrolysis

Question 23.
Intermediate compound formation theory explains.
a) Homogeneous catalysis
b) Heterogeneous catalysis
c) Autocatalysis
d) Negative catalysis
Answer:
a) Homogeneous catalysis

Question 24.
In intermediate compound formation, the intermediate complex formed has
a) More activation energy than uncatalysed complex
b) Less activation energy than uncatalysed complex
c) Less kinetic energy than the reactants
d) Less kinetic energy than the products
Answer:
b) Less activation energy than uncatalysed complex

Question 25.
Adsorption theory explains
a) Homogeneous catalysis
b) Heterogeneous catalysis
c) Autocatalysis
d) Negative catalysis
Answer:
b) Heterogeneous catalysis

Question 26.
The catalytic activity of a catalyst is increased by a promoter by
a) Increasing the number of active centres
b) Decreasing the number of active centres.
c) Blocking the number of active centres
d) Desorbing the active centres
Answer:
a) Increasing the number of active centres

Question 27.
Enzymes are often present as
a) Crystalloids
b) Suspension
c) Colloids
d) True solutions
Answer:
c) Colloids

Question 28.
In the conversion of 1 – chloro octane into 1 – cyano octane, tetra alkyl ammonium cation acts as a
a) Enzyme catalyst
b) Phase transfer catalyst
c) Zeolite catalyst
d) Nano catalyst
Answer:
b) Phase transfer catalyst

Question 29.
Which of the following is incorrect
a) Enzymes can be inhibited (poisoned)
b) Calalytic activity of enzymes is decreased by coenzvmes.
c) Enzyme catalysis is highly specific in nature
d) the rate of Enzyme catalysed reactions varies with the pH of the system.
Answer:
b) Calalytic activity of enzymes is decreased by coenzymes.

Question 30.
Nano catalysts can act as
a) Homogeneous catalysts
b) Heterogeneous catalysts
c) Phase transfer catalysts
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Question 31.
Size of colloidal particle is
a) 1 – 200 Å
b) 1 – 200 nm
c) 1 – 200 pm
d) 1 – 200 pm
Answer:
b) 1 – 200 nm

Question 32.
In hydrosols the dispersion medium is
a) Benzene
b) Alcohol
c) Water
d) Ether
Answer:
c) Water

Question 33.
In lyophillic colloids which is true?
a) Definite attractive force exists between the dispersion medium and dispersed phase
b) More stable
c) Reversible sols
d) All the above
Answer:
d) All the above

Question 34.
In lyophobic colloids which is not true?
a) No attractive force exists between the dispersion medium and dispersed phase
b) Less stable
c) Can be produced again
d) Precipitated readily
Answer:
c) Can be produced again

Question 35.
Which among the following is a liquid aerosol?
a) Smoke
b) Fog
c) Shaving cream
d) Froth
Answer:
b) Fog

Question 36.
Which one of the following is negatively charged colloid?
a) arsenic sulphide
b) Ferric hydroxide
c) Haemoglobin
d) Basic dyes
Answer:
a) arsenic sulphide

Question 37.
Butter is a colloid of
In butter the dispersed phase and the
dispersion medium are respectively [222EE3
a) Solid, gas
b) Liquid, solid
c) Solid, liquid
d) Gas, solid
Answer:
c) Solid, liquid

Question 38.
In butter the dispersed phase and the dispersion medium are respectively_______
a) Solid in liquid
b) Liquid in solid
c) Liquid in liquid
d) Liquid in gas
Answer:
b) Liquid in solid

Question 39.
Colloidal graphite can be prepared by
a) Peptisation
b) Mechanical dispersion
c) Ultrasonic dispersion
d) Double decomposition
Answer:
b) Mechanical dispersion

Question 40.
Electro dispersion method is used to prepare the colloidal solution of
a) Copper
b) Silver
c) Gold
d) All the above
Answer:
d) All the above

Question 40.
Arsenic sulphide sol is prepared by the reaction. AS₂O₃ + 3H₂S → As₂S₃ + 3H₂O. This method is known as
a) Hydrolysis
b) Double decomposition
c) Oxidation
d) Reduction
Answer:
b) Double decomposition

Question 42.
Conversion of a colloid into a precipitate is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
c) Coagulation

Question 43.
Conversion of a precipitate into a colloid is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
a) Peptisation

Question 43.
The removal of electrolytic impurities from a colloidal solution is known as
a) Peptisation
b) Dialysis
c) Coagulation
d) Electrophoresis
Answer:
b) Dialysis

Question 44.
The movement of dispersed phase under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
b) Dialysis
d) Electrophoresis
Answer:
b) Dialysis

Question 45.
The movement of dispersed phase under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
c) Electrodialysis
d) Ultra filteration
Answer:
b) Electrophoresis

Question 46.
The movement of dispersion medium under the influence of electric current is known as
a) Electro osmosis
b) Electrophoresis
c) Electrodialysis
d) Ultra filteration
Answer:
a) Electro osmosis

Question 47.
In ultra filteration, ultrafilters are made by using
a) Collodion
b) Cellophane
c) Both (a) and (b)
d) None of the above
Answer:
c) Both (a) & (b)

Question 48.
Collodion is 4% solution of ……………… in a mixture of alcohol and water.
a) Cellulose acetate
c) Chioro cellulose
b) Cellulose sulphate
cl) Nitrocellulose
Answer:
d) Nitrocellulose

Question 49.
Blue colour of the sky. in nature is due to
a) Browniart movement
b) Tyndall effect
c) Both (a) and (b)
d) None of thé above
Answer:
b) Tyndall effect

Question 50.
Scattering of light by colloidal particles is known as
a) Brownian movement
b) Tyndall effect :
c) Electrophoresis
d) Electro osmosis
Answer:
b) Tyndall effect

II. Match the following

Question 1.

Answer:
i) Arsenic in the decomposition of arsine
i) Fe/Pd
ii) Hydrated aluminosilicates
iii) As₂O₃ in contact process
iv) Al₂O₃ in Haber’s process
v) Urease

Question 2.

Answers :
i) Iron
ii) Platinum
iii) Nitric oxide
iv) Cupric chloride
v) Ferric oxide

Question 3.

Answer:
i) Dust
ii) Milk
iii) Paints
iv) Bread
v) Alloys

III. Assertion and Reasoning

Question 1.
Assertion (A) : Physical adsorption occurs at low temperatures.
Reason (R) : The forces of attraction between the adsorbent and adsorbate are weak and heat of adsorption is low
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct.
Answer:
a) Both A and R are correct, R explains A

Question 2.
Assertion (A) : In the reaction CH3COOC2H5 + H20 -> CH3COOH + C2H5OH is an example for autocatalysis.
Reason (R) : One of the product ethanol acts as a catalyst.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct.
Answer:
c) A is correct but R is wrong.
Correct Reason: One of the product acetic acid acts as a catalyst.

Question 3.
Assertion (A) : 1- Chioro octane is converted into 1 – Cyano octane in I or 2 hours by reacting with sodium cyanide in presence of tetra alkyl ammonium chloride.
Reason (R) : The tetra alkyl ammonium cation transports CN^– from the aqueous phase to the
organic phase using its hydrophilic end.
a) Both A and R are correct, R explains A
h) Both A and R are correct, R does do not explain A
c) A is correct hut R is wrong
d) A is wrong but R is correct.
Answer:
a) Both A and R are correct, R explains A

Question 4.
Assertion (A): Measurements of osmotic pressure is used to find the molecular weight of colloidal particle.
Reason (R): Colloidal solutions show the kinetic property,
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct.
Answer:
b) Both A and R are correct, R does not explain A Correct Reason: Colloidal solutions show colligative properties.

IV. Choose the correct statement

Question 1.
i) Chemisorption is instantaneous
ii) Chemisorption first increases and then decreases with temperature
iii) In chemisorption the heat of adsorption is high
iv) Chemisorption is independent of the surface area of the adsorbent
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
Chemisorption is slow,
Chemisorption depends on the surface area of the adsorbent.

Question 2.
i) A catalyst can initiate a reaction
ii) A solid catalyst will be more effective if it is taken in bulk
A catalyst does not affect the position of equilibrium and the value of the equilibrium constant
A catalyst is highly effective at a particular temperature called optimum temperature,
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)

Correct statement:
A catalyst cannot initiate a reaction,
A solid catalyst will be more effective if it is taken in a finely divided form.

Question 3.
i) When light passes through colloidal solution it is scattered in all directions.
ii) When a colloidal solution is observed through ultramicroscope, they showed a random, zigzag, ceaseless motion.
iii) Haemoglobin is a negatively charged colloid
iv) A sol is not electrically neutral.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
a) (i) & (ii)

Correct statement: iii) Haemoglobin is a positively charged colloid, iv) A sol is electrically neutral

Question 4.
i) The flocculation and setting down of the sol particles is called coagulation.
ii) The higher the flocculation value greater will be the precipitation.
iii) Lyophilic sols are precipitated readily even with small amount of electrolytes.
iv) When boiled due to increased collisions, the sol particles combine and settle down.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)

Correct statement:
ii) The smaller the flocculation value greater will be the precipitation,
iii) Lyophobic sols are precipitated readily even with small amount of electrolytes

V. Choose the incorrect statement

Question 1.
i) Physisorption decreases with an increase in pressure
ii) Physisorption decreases with an increase in temperature
iii) Activation energy of physisorption is significant
iv) No transfer of electrons occur in physisorption
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (iii) & (iv)
Answer:
c) (i) & (iii)

Correct statement:
i) Physisorption increases with increase in pressure,
iii) Activation energy of physisorption is insignificant.

Question 2.
i) Enzyme catalysis is highly specific in nature
ii) The rate of enzyme catalysed reactions does not vary with pH of the system
iii) Catalytic activity of enzymes is decreased by coenzymes.
iv) Enzyme catalysed reaction has maximum rate at optimum temperature.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)

Correct statement:
ii) The rate of enzyme catalysed reactions varies with pH of the system,
iii) Catalytic activity of enzymes is increased by coenzymes.

Question 3.
i) In zeolites the active sites namely protons are lying inside their pores.
ii) Reactions occur only outside the pores of zeolites.
iii) Bulkier reactant molecules are prevented from reaching the active sites within- the zeolite crystal.
iv) If the transition state of a reaction is large compared to the pore size of the zeolite, then product will be formed.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (ii) & (iv)
Answer:
d) (ii) & (iv)
Correct statement:
ii) Reactions occur only inside the pores of zeolites,
iv) If the transition state of a reaction is large compared to the pore size of the zeolite, then no product will be formed.

Question 4.
i) Colloidal solutions are heterogeneous in nature having two distinct phases.
ii) Colloidal solutions are unstable and they are affected by gravity.
iii) When the colloidal solution is dilute coagulation occurs.
iv) Unlike true solution, colloids diffuse less readily through membranes.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
ii) Colloidal solutions are quite stable and they are not affected by gravity,
iii) When the colloidal solution is dilute coagulation does not occur

VI. Two Mark Questions

Question 1.
What is adsorption?
Answer:

• Adsorption is a phenomenon of concentration of a substance on the surface of a liquid or gas.
• Adsorption is a surface phenomenon.

Question 2.
What is known as interface?
Answer:
The surface of separation of the two phases where the concentration of adsorbed molecule is high is known as the interface.

Question 3.
What is meant by positive adsorption and negative adsorption?
Answer:
If the concentration of a substance in the interface is high in an adsorption, then it is called positive adsorption. If it is less, then it is called negative adsorption.

Question 4.
What is desorption?
Answer:
The process of removing an adsorbed substance from the surface is called desorption.

Question 5.
What is meant by adsorbent and adsorbate?
Answer:
Absorbent is the material on which adsorption takes place.
Adsorbed substance is called an adsorbate.

Question 6.
Give some examples for adsorbates.
Answer:

• Gaseous molecules like He, Ne, O₂, SO₂ and NH₃
• Solutions of NaCl, KCl

Question 7.
Give some examples for adsorbents.
Answer:
Silica Gel
Metals like Ni, Cu, Pi. Ag and Pd
Certain colloids.

Question 8.
What are the limitations of Freundlich isotherm?
Answer:

• The equation is purely empirical
• Valid over a limited pressure range.
• Constants k and n values vary with temperature and for this no theoretical explanations were given.

Question 9.
How is adsorption principle used in the softening of hard water?
Answer:

• Permutit adsorbs Ca²⁺ and Mg²⁺ ions present in hard water on its surface.
• An ion exchange occurs on its surface. Na₂Al₂Si₄O₁₂ + CaCl₂ → CaAl₂ Si₄ O₁₂ + 2NaCl
• Exhausted permutit is regenerated by adding a solution of common salt.
  CaAl₂ Si₄ O₁₂ + 2NaCl → Na₂Al₂Si₄O₁₂ + CaCl₂

Question 10.
How is ion exchange resins work on the principle of adsorption in demineralising water?
Answer:

• The process is carried out by passing water through two columns of cation and anion exchange resins.
• Cation exchange:
  
• Anion exchange:
  

Question 11.
Define a catalyst.
Answer:
A catalyst is defined as a substance which alters the rate of chemical reaction without itself undergoing chemical change.

Question 12.
What is meant by positive catalysis and negative catalysis?
Answer:

• In positive catalysis the rate of a reaction is increased by the presence of a catalyst.
• In negative catalysis the rate of a reaction is decreased by the presence of a catalyst.

Question 13.
What are promoters?
Answer:

• The substance which increases the activity of a catalyst in a reaction is called promoter.
• (Eg-) In Haber’s process of synthesis of NH₃, MO is the promoter of Fe catalyst.

Question 14.
What are the limitations of intermediate compound formation theory.
Answer:

• Fails to explain the action of catalytic poison and promoters.
• Unable to explain the mechanism of heterogeneous catalysis.

Question 15.
What are active centres?
Answer:

• Surface of a catalyst is not smooth.
• It bears steps, cracks and corners.
• The atoms on such locations are coordinatively unsaturated.
• So they have much residual force of attraction.
• Such sites are called active centres.

Question 16.
What is a colloidal solution?
Answer:
When the diameter of the particles of a substance dispersed in a solvent ranges from 1 – 200 nm, the system is called a colloidal solution.

Question 17.
What is meant by dispersion medium and dispersed phase?
Answer:
In a colloid, the substance present in a larger amount is called dispersion medium and the substance present in less amount is called the dispersed phase.

Question 18.
What is the flocculation value?
Answer:
The precipitation power of electrolyte is determined by finding the minimum concentration(millimoles/lit ) required to cause precipitation of a sol in 2 hours. This value is called flocculation value.
The smaller the flocculation value greater will be precipitation.

Question 19.
Name the dispersion medium present in hydrosol, alcosol and benzosol.
Answer:

Question 20.
How is a colloid prepared by exchange of solvent ?
Answer:

• Colloidal solution of phosphorous or sulphur is obtained by preparing the solutions in alcohol and pouring them into water.
• As they are insoluble in water, they form colloidal solution P in alcohol + water —> P Sol

Question 21.
What is Tyndall effect?
Answer:
The scattering of light by colloidal particles is called Tyndall effect.

Qu

estion 22.
What is Brownian movement?
Answer:

• The random, zigzag, ceaseless motion of colloidal particles in dispersion medium is known as Brownian movement.
• This is due to the continuous bombardment of dispersed phase bv the molecules of dispersion medium.

Question 23.
What is the significance of Brownian movement ?
Answer:

• To calculate Avogadro number.
• To confirm kinetic theory which considers the ceaseless rapid movement of molecules that increases with increase in temperature.
• To understand the stability of colloids.
• As the particles are in continuous rapid movement they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted by force of gravity.

Question 24.
Define ‘gold number’
Answer:

• Gold number is defined as the number of milligrams of hydrophilic colloid that will just prevent the precipitation of 10 ml of gold sol on the addition of 1 ml of 10% NaCl solution.
• Smaller the gold number greater the protective power.

Question 25.
What are emulsions?
Write their types.
Answer:

• Emulsions are colloidal solution in which a liquid is dispersed in another liquid.
• Two types of emulsions are
  (i) Oil in water (O/W)
  (ii) Water in oil (W/ O)

Question 26.
What is emulsification?
Answer:
The process of preparation of emulsion by the dispersal of one liquid in another liquid is called emulsification.

Question 27.
What is de emulsification?
Answer:
The process of separation of an emulsion into two separate layers is called de emulsification.

Question 28.
Write the uses of colloids in medicine?
Answer:

• Antibodies such as penicillin and streptomycin are produced in colloidal form for suitable injections.
• Colloidal gold and colloidal calcium are used as tonics.
• Milk of magnesia is used for stomach troubles.
• Silver sol protected by gelatine known as Argyrol is used as eye lotion.

Question 29.
Mention the shapes of the following colloidal particles,
i) AS₂S₃
ii) Blue gold sol
iii) Tungstic acid sol
Answer:
i) AS₂S₃ -Spherical
ii) Blue gold sol – Disc or plate like
iii) Tungstic acid sol – Rod like

VII. Three Mark Questions

Question 1.
Write the characteristics of adsorption
Answer:

• Adsorption can occur in all interfacial surfaces (ie) between gas-solid, liquid-solid, liquid-liquid, solid-solid, gas-liquid
• Adsorption is always accompanied by decrease in free energy. When ΔG = 0, the equilibrium is attained.
• Adsorption is a spontaneous process.
• Adsorption is accompanied by decrease in randomness (ie) decrease in entropy.
• Adsorption is exothermic.
• Adsorption is a quick process.

Question 2.
Write a note on Freundlich isotherm.
Answer:
A plot between the amount of adsorbate adsorbed and pressure or concentration of adsorhate at constant temperature is called adsorption isotherm.

According to Freundlich x/m = kp^1/n
x = amount of adsorbate
m = mass of adsorbent in gram
p = pressure
k, n = constants, value of n is always less than unity.

This equation is applicable for adsorption of gases on solid surfaces.
For adsorption in solutions of concentration C, the equation is x/m = kC^1/n
These equations quantitatively predict the effect of pressure or concentration on the adsorption of gases or liquids at constant temperature
Taking log

Question 3.
Write a note on (i) auto catalysis , (ii) negative catalysis ?
Answer:
Auto catalysis:
When one of the products formed acts as a catalyst to the reaction it is called as auto catalysis and that product is-called as auto catalyst.
(Eg.) CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂ + C₂H₅OH
Acetic acid acts as the auto catalyst.

Negative catalysis:
When certain substances decreases the rate of the reaction, it is called negative catalysis and the substance is called negative catalyst.
(Eg.) 2H₂O₂ → 2H₂O + O₂
Glycerol acts as a negative catalyst in the above reaction.

Question 4.
What is the role of adsorption in the heterogeneous catalysis.
Answer:
In heterogenous catalysis, the reactants are converted into products only when the reactants are adsorbed on the catalyst surface to form an activated complex which subsequently decomposes to give the product. Hence the role of adsorption is to form activated complex.

Question 5.
Give some examples for enzyme catalysis.
Answer:

Question 6.
Write a note on nano catalysis.
Answer:

• Metallic nano particles, metal oxides are used as catalysts.
• Nano catalysts can act as both homogeneous as well as heterogeneous catalysts.
• Like homogeneous catalysts they give 100% selective transformations and excellent yield and show extremely high activity.
• Like heterogeneous catalysts, they can be recovered and recycled.
• Nano catalysts are actually soluble heterogeneous catalysts.

Question 7.
How can you identify the two types of emulsion.
Answer:

Question 8.
Name some of the de emulsification techniques.
Answer:

• Distilling one component.
• Adding an electrolyte to destroy the charge. Destroying the emulsifier using chemical methods.
• Using solvent extraction to remove one component.
• By freezing one of the components.
• By applying centrifugal force.
• Adding dehydrating agents for water in oil (W/O) type.
• Using ultrasonic waves.
• Heating at high pressure.

VIII. Five Mark Questions

Question 1.
What are the general characteristics of catalysts.
Answer:

• Only a small quantity of catalyst is needed for a reaction.
• There may be some physical changes, but the catalyst remains unchanged’ in mass and its chemical composition.
• A catalyst cannot initiate a reaction, only it speeds up a slow reaction.
• A solid catalyst is more effective in a finely divided form.
• Catalysts are specific in nature.
• A catalyst does not affect the position of equilibrium and the value of equilibrium constant but helps the quick attainment of equilibrium.
• A catalyst is highly effective at a particular temperature called as optimum temperature.
• Presence of a catalyst does not change the nature of products.

Question 2.
What are the special characteristics shown by enzyme catalysed reaction?
Answer:
Effective and efficient conversion An enzyme may transform a million molecules of reactant in a minute. (Eg.) 2H₂O₂ → 2H₂O + O₂

Enzyme catalysis has maximum rate at optimum temperature. The temperature at which the enzymic activity is high or maximum is called as optimum temperature.
Enzymes involved in human body have an optimum temperature 37°C.
The rate of enzyme catalysis varies with pH of the system. The pH at which the rate is maximum is called the optimum pH.
Enzymes can be inhibited or poisoned. Penicillin inhibits the action of bacteria and used for curing diseases like pneumonia, dysentery, cholera.
A small non protein (vitamin) called a coenzyme promotes the catalytic activity of enzyme.

Question 3.
Explain phase transfer catalysis.
Answer:

• If the reactants of a reaction are present in two different solvents which are immiscible, the reaction between them is very slow.
• As the solvents form separate phases, the reactants have to migrate across the boundary to react.
• But migration across the boundary is not easy.
• For such a situation a third solvent miscible with both is added.
• So the phase boundary is eliminated and the reactants mix freely and react fast.
• But for large scale production of any product, use of a third solvent is not convenient as it may be expensive.
• Phase transfer catalysis gives a simple solution for such problems as it avoids the use of solvents.
• Phase transfer catalyst facilitate the transport of a reactant in one solvent to the other solvent where the second reactant is present.
• As the reactants are brought together, they rapidly react and form the product.
  R – Cl + NaCN → R-CN + NaCl
  Organic phase aqueous phase Organic phase aqueous phase
• RC1 = 1 – chloro octane, RCN = 1 – cyano octane
  1- cyano octane is not obtained by directly heating organic phase, 1 – chloro octane ^ with aqueous phase sodium cyanide for^ ’ several days.
• If a small amount of tetra alkyl ammonium chloride is added, 1 – cyano octane occurs in about 100% yield after 1 or 2 hours.
• In this reaction, the tetra alkyl ammonium cation containing hydrophobic and hydrophilic ends, transports CN^– from the aqueous phase to the organic phase using its hydrophilic end.
  
• Both in organic phase Organic phase
  It moves to aqueous phase, releases Cl— again picks up CN^– and transports it.
• So phase transfer catalyst speeds up the reaction by transporting one reactant from one phase to another.

Question 4.
Explain dispersion methods of preparing colloids
Answer:

Mechanical Dispersion:
The solid is ground to colloidal dimension using a colloid mill consists of two metal plates rotating in opposite direction at very high speed of nearlv 7000 revolution / minute.
Required colloidal si/.e is obtained bv adjusting the distance between the two plates.
(Eg.) Colloids of ink and graphite.

ii) Electro Dispersion :
An electric arc is struck between electrodes dispersed in water surrounded by ice.
When a current of l amp/100 v is passed, an arc produced forms vapours of metal which immediate!) condense to form colloidal solution
Alkali hvdroxide is added as an stabilising agent.
(Eg.) Colloids of metals like copper, silver, gold, platinum
Svedberg modified this method for the prepartion of non-aqueous inflammable liquids like pentane, ether and benzene using high frequency alternating current which prevents the decomposition of the liquid.

iii) Ultrasonic Dispersion:
Sound wraves of frequency more than 20 KHz transform coarse suspension to colloids. Claus obtained mercury sol by subjecting mercury to sufficiently high frequency ultrasonic vibrations.
Ultrasonic vibrations produced by generator spread through the oil and transfer the vibration to the vessel with mercury in water.

iv) Peptisation:
Conversion of a precipitate into colloid by the addition of electrolyte is called peptisation.
Added electrolyte is called peptising or dispersing agent.

Question 5.
Explain condensation methods of preparing colloids.
Answer:
i) Oxidation:
Sols of non metals are prepared by this method
HIO₃ + 5HI → 3H₂O + I₂ (sol)

ii) Reduction:
Gold sol is prepared by reducing auric chloride using organic reagent formaldehyde.
2AUCl₃ + 3HCHO + 3H₂ → 2Au (sol) + 6HCl + 3HCOOH

iii) Hydrolysis :
Sols of metal hydroxides are prepared by this method.
FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl

Double decomposition:
Water insoluble sols are prepared by this method. Yellow coloured arsenic sulphide sol is prepared by passing hydrogen sulphide gas through arsenic oxide.
AS₂O₃+ 3H2S → AS₂S₃ + 3H₂O

vi) Decomposition:

• When a few drops of an acid is added to sodium thio sulphate, it decomposes to form insoluble free sulphur.
• Insoluble free sulphur accumulates into small clusters which impart various colours blue, yellow and even red to the system.
• This colour change is due to their growth within the size of colloidal dimensions.
  

Question 6.
Explain various methods of purification of colloids.

Answer:
Dialysis:

• Colloidal solution is taken in a bag made up of semipermeable membrane.
• It is suspended in a trough of flowing water.
• Electrolytes diffuse out of the membrane and they are carried away by water.

Electro dialysis :

• The presence of electric field increases the speed of removal of electrolytes from colloidal solution.
• Colloidal solution with electrolytic impurity is placed between two dialysing membranes enclosed into two compartments filled with water.
• When current is passed, the impurities pass into water compartment and get removed periodically.
• This process is faster than dialysis, as the rate of diffusion of electrolytes is increased by the application of electricity.

iii) Ultra filtration:

• The pores of ordinary filter papers allow the passage of colloidal particles.
• In ultra filtration the membranes are made by using colloidion, cellophane or visiking.
• When a colloidal solution is filtered using such a filter, colloidial particles are separated on the filter and the impurities are removed as washings.
• This process is quickened by the application of pressure.
• Collodion is 4% solution of nitrocellulose in a mixture of alcohol and water.
• This method of separation of sol particles from electrolyte by filtration through an ultrafilter is called ultrafiltration.

Question 7.
Write a note on Helmholtz double layer.

• The surface of colloidal particle adsorbs one type of ion due to preferential adsorption
• This layer attracts the oppositely charged ions in the medium.
• Hence at the boundary separating the two, electrical double layers are set up.
• This is called as Helmholtz electrical double layer.
• As the particles nearby are having similar charges, they cannot come close and condense.
• This explains the stability of a colloid.

Question 8.
Write a note on electrophoresis.
Answer:

• The migration of sol particles towards an electrode under the influence of electric field is called electrophoresis or cataphoresis.
• When electric potential is applied across two platinum electrodes dipped in a hydrophilic sol, the dispersed particles move toward one or other electrode.
• If the sol particles migrate to the cathode then they possess positive charges.
• If the sol particles migrate to the anode then thev have negative charges.
• Thus from the direction of migration of sol particles, the charge on them can be determined.
• Hence electrophoresis is used for the detection of presence of charges on the sol practicles.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.3

Question 1.
The subscription of a magazine sends out a letter to a large mailing list inviting subscriptions for the magazine. Some of the people receiving this letter already subscribe to the magazine while others do not. From this mailing list, 45% of those who already subscribe will subscribe again while 30% of those who do not now subscribe will subscribe. On the last letter, it was found that 40% of those receiving it ordered a subscription. What percent of those receiving the current letter can be expected to order a subscription?
Solution:
Transition Probability Matrix

S = 36%; F = 64%
∴ 36% of those receiving the current letter can be expected to order a subscription

Question 2.
Anew transit system has just gone into operation in Chennai. Of those who use the transit system this year, 30% will switch over to using metro train next year and 70% will continue to use he transit system. Of those who use metro train this year and 70% will continue to use metro train next year and 30% will switch over to transit system. Suppose the population of Chennai city remains constant and that 60% of the commuters use the transit system and 40% of the commuters use metro train this year.
(i) What percent of commuters will be using the transit system after one year?
(ii) What percent of commuters will be using the transit system in the one run?
Solution:
Transition Probability Matrix

S = 54%; C = 46%
Equilibrium will be reached in the long run
At equilibrium we must have
(S C) T = (S C) where S + C = 1
(S C) = \(\left(\begin{array}{ll}
0.7 & 0.3 \\
0.3 & 0.7
\end{array}\right)\) = (S C)
0.7S + 0.3C = S
0.7S + 0.3 (1 – S) = S
0.7S + 0.3 – 0.3S = S
0.4S + 0.3 = S
0.3 = S – 0.4S ⇒ 0.6S = 0.3
S = \(\frac { 0.3 }{ 0.6 }\) = 0.50
∴ 50% of the commuters will be transit system the long run.

Question 3.
Two types of soaps A and B are in the market. Their present market shares are 15% for A and 85% for B. Of those who bought A the previous year, 65% contionues to buy it again while 35% switch over to B. Of those who bought B the previous year, 55% buy it again and 45% switch over to A. Find their market shares after one year and when is the equilibrium reached?
Solution:
Transition Probability Matrix

Equilibrium will be reached in the long run
At equilibrium we must have
(A B) T = (A, B) where A + B = 1
(A B) = \(\left(\begin{array}{ll}
0.65 & 0.35 \\
0.45 & 0.55
\end{array}\right)\) = (A B)
0.65 A + 0.45 B = A
0.65 A + 0.45 (1 – A) = A
0.65 A + 0.45 – 0.45 A = A
0.20 A + 0.45 = A
A – 0.20 A = 0.45
0.80 A = 0.45
A = \(\frac { 0.45 }{ 0.80 }\) = 0.5625 and B = 04375
In the long sum {∴ B = 1 – A}
∴ A = 56.25 % and B = 43.75%

Question 4.
Two products A and B currently share the market with shares of 50% and 50% each respectively. Each week some brand switching takes place. Of those who bought A the previous week, 60% buy it again whereas 40% switch over to B. Of those who bought B the previous week, 80% buy it again whereas 20% switch over to A. Find their shares after one week and after two weeks. If the price war continues, when is the equilibrium reached?
Solution:
Transition Probability Matrix

Equilibrium will be reached in the long run.
At equilibrium, we must have
(A B) T = (A B) where A + B = 1
(A B) = \(\left(\begin{array}{ll}
0.60 & 0.40 \\
0.20 & 0.80
\end{array}\right)\) = (A B)
0.60 A + 0.20 B = A
0.60 A+ 0.20 (1 – A) = A
0.60 A + 0.20 – 0.20 A = A
0.40 A + 0.20 = A
0.20 = A – 0.40 A
0.60 A = 0.20
A = \(\frac { 0.20 }{ 0.60 }\) = 0.33 and B = 0.67
{∴ B = 1 – A}
In the long run
∴ A = 33% and B = 67%

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.2

Question 1.
Solve the following . equations by using Cramer’s rule
(i) 2x + 3y = 7; 3x + 5y = 9
(ii) 5x + 3v = 17; 3x + 7y = 31
(iii) 2x + y – z = 3; x + y + z – 1; x – 2y – 3z = 4
(iv) x + y + z = 6; 2x + 3y – z = 56; x – 2y – 3z = -7
Solution:
The equations are
2x + 3y = 7
3x + 5y = 9
Here Δ = \(\left|\begin{array}{ll}
2 & 3 \\
3 & 5
\end{array}\right|\) = 10 – 9 = 1
≠ 0
∴ We can apply Cramer’s Rule

∴ x = 8; y = -3

(ii) The equations are
5x + 3y = 17
3x + 7y = 31
Here Δ = \(\left|\begin{array}{ll}
5 & 3 \\
3 & 7
\end{array}\right|\) = 35 – 9 = 26
≠ 0
We can apply Cramer’s Rule

∴ x = 1; y = 4

(iii) The equations are
2x + y – z = 3
x + y + z = 1
x – 2y – 3z = 4
Here Δ = \(\left|\begin{array}{ccc}
2 & 1 & -1 \\
1 & 1 & 1 \\
1 & -2 & -3
\end{array}\right|\)
= 2 (-3 + 2) – 1 (-3 – 1) – 1 (-2 – 1)
= 2 (-1) -1 (-4) -1 (-3)
= -2 + 4 + 3
Δ = 5 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistant and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
3 & 1 & -1 \\
1 & 1 & 1 \\
4 & -2 & -3
\end{array}\right|\)
= 3 (-3 + 2) – 1 (-3 – 4) – 1 (-2 – 4)
= 3 (-1) -1 (-7) -1 ( 6)
= -3 + 7 + 6 = 10
Δ_y = \(\left|\begin{array}{ccc}
2 & 3 & -1 \\
1 & 1 & 1 \\
1 & 4 & -3
\end{array}\right|\)
= 2 (- 3 – 4) -3 (-3 – 1) -1 (4 – 1)
= 2 (-7) – 3 (-4) – 1 (3)
= -14 + 12 – 3
= -5
Δ_z = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
1 & 1 & 1 \\
1 & -2 & 4
\end{array}\right|\)
= 2 (4 + 2) – 1 (4 – 1) + 3 (- 2 – 1)
= 2 (6) – 1(3) + 3 (-3)
= 12 – 3 – 9 = 0
∴ By Cramer’s rule

∴ (x, y, z) = (2, -1, 0)

(iv) The equations are
x + y + z = 6
2x + 3y – z = 5
6x – 2y – 3z = -7
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & -1 \\
6 & -2 & -3
\end{array}\right|\)
= 1 (-9 – 2) -1 (-6 + 6) + 1 (-4 – 18)
= 1 (-11) – 1 (0) +1 (-22)
= -11 – 22
= -33 ≠ 0
∴ We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Now,
Δ_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
5 & 3 & -1 \\
-7 & -2 & -3
\end{array}\right|\)
= 6 (- 9 – 2) – 1 (- 15 – 7) + 1 (- 10 + 21)
= 6 (-11) – 1 (-22)+ 1 (11)
= -66 + 22 + 11
= -33
Δ_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
2 & 5 & -1 \\
6 & -7 & -3
\end{array}\right|\)
= 1 (-15 – 7) – 6 (- 6 + 6) + 1 (- 14 – 30)
= 1 (-22) – 6 (0) + 1 (-44)
= -66
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
2 & 3 & 5 \\
6 & -2 & -7
\end{array}\right|\)
= 1 (- 21 + 10) – 1 (- 14 – 30) + 6 (- 4 – 18)
= 1 (-11) – 1 (-44) + 6 (-22)
= -11 + 44 – 132
= -99
∴ By Cramer’s rule

(x, y, z) = (1, 2, 3)

(v) The equations are
x + 4y + 3z = 2
2x – 6y + 6z = -3
5x – 2y + 3z = -5
Here Δ = \(\left|\begin{array}{ccc}
1 & 4 & 3 \\
2 & -6 & 6 \\
5 & -2 & 3
\end{array}\right|\)
= 1 (-18 + 12) -4 (6 -30) + 3 (-4 + 30)
= 1 (-6) – 4 (-24) + 3 (26)
= -6 + 96 + 78
= 168 ≠ 0
We can apply Cramer’s Rule and the system is consistent and it has unique solution.
Δ_x = \(\left|\begin{array}{ccc}
2 & 4 & 3 \\
-3 & -6 & 6 \\
-5 & -2 & 3
\end{array}\right|\)
= 2 (-18 + 12) – 4 (-9 + 30) + 3 (6 – 30)
= 2 (-6) -4 (21) + 3 (-24)
= -12 – 84 – 72
= -168
Δ_y = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & -3 & 6 \\
5 & -5 & 3
\end{array}\right|\)
= 1 (-9 + 30) – 2 (6 – 30) + 3 (-10 + 15)
= 1(21) – 2 (-24) + 3 (5)
= 21 + 48 + 15
= 84
Δ_z = \(\left|\begin{array}{ccc}
1 & 4 & 2 \\
2 & -6 & -3 \\
5 & -2 & -5
\end{array}\right|\)
= 1 (30 – 6) – 4 (-10 +15) + 2 (-4 + 30)
= 1 (24) – 4 (5) + 2 (26)
= 24 – 20 + 52
= 56
∴ By Cramer’s rule

(x, y, z) = (-1, \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\))

Question 2.
A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is Rs 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is Rs 56. What is the cost per unit of labour and capital? (Use determinant method).
Solution:
Let ‘x’ be the cost per unit of labour
Let ‘y’ be the capital
∴ 3x + 2y = 62
4x + y = 56

∴ The cost per unit of labour is Rs 10 and Cost per unit of capital is Rs 16

Question 3.
A total of Rs 8,600 was invested in two accounts. One account earned 4\(\frac { 3 }{ 4 }\)% annual interest and the other earned 6\(\frac { 1 }{ 2 }\)% annual interest. If the total interest for one year was Rs 431.25, how much was invested in each accout? (Use determinant method)
Solution:
Let the two investments be x and y
Given that
x + y = 8600 ……. (1)

Multiply by 4 (both sides)
19x + 26y = 172500 ……….. (2)
Here Δ = \(\left|\begin{array}{ll}
1 & 1 \\
19 & 26
\end{array}\right|\) = 26 – 19 = 7 ≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ll}
8600 & 1 \\
172500 & 26
\end{array}\right|\)
= 223600 – 172500
= 51100
Δ_y = \(\left|\begin{array}{ll}
1 & 8600 \\
19 & 172500
\end{array}\right|\)
= 172500 – 163400
= 9100
By Cramer’s Rule
x = \(\frac { Δ_x }{ Δ }\) = \(\frac { 51100 }{ 7 }\) = 7300
y = \(\frac { Δ_y }{ Δ }\) = \(\frac { 9100 }{ 7 }\) = 1300
∴ Amount invested at 4\(\frac { 3 }{ 4 }\)% is Rs 7,300 and
Amount invested at 6\(\frac { 1 }{2 }\)% is Rs 1,300

Question 4.
At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Bentia spent Rs 780 and Rs 560 during the month of May

Find the hourly charges for the two games (rides). (Use determinant method).
Solution:
let the hourly charges for horse riding be x and let the hourly charges for Quad Bikes riding be y.
For Keren
3x + 4y = 780 ……… (1)
For Benita
2x + 3y = 560 ……… (2)
Here Δ = \(\left|\begin{array}{ll}
3 & 4 \\
2 & 3
\end{array}\right|\) = 9 – 8 = 1 ≠ 0
∴ We can apply Cramer’s Rule

∴ The hourly charges for horse riding is Rs 100 and Quad Bikes riding is Rs 120

Question 5.
In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides the information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Find the weights assigned to the three varieties by using Cramer’s Rule.
Solution:
Let x, y and z be are consumption of three commodities A, B and C respectively.
Given that
x + 2y + 3z = 11 ……… (1)
2x + 4y + 5z = 21 ……… (2)
3x + 5y + 6z = 27 ……… (2)
Here Δ = \(\left|\begin{array}{ccc}
1 & 2 & 3 \\
2 & 4 & 5 \\
3 & 5 & 6
\end{array}\right|\)
= 1 (24 – 25) – 2 (12 – 15) + 3 (10- 12)
= 1 ( 1) 2 ( 3) + 3 (- 2)
= -1 + 6 – 6 = -1 ≠ 0
∴ We can apply Cramer’s Rule
Now Δ_x = \(\left|\begin{array}{ccc}
11 & 2 & 3 \\
21 & 4 & 5 \\
27 & 5 & 6
\end{array}\right|\)
= 11 (24 – 25) -2 (126 – 135) + 3 (105 – 108)
= 11(-1) -2 (-9)+ 3 (-3)
= -11 + 18 – 9
= -2
Δ_y = \(\left|\begin{array}{ccc}
1 & 11 & 3 \\
2 & 21 & 5 \\
3 & 27 & 6
\end{array}\right|\)
= 1 (126 – 135) -11 (12 – 15) + 3 (54 – 63)
= 1 (-9) -11 (-3) + 3 (-9)
= -9 + 33 – 27
= -3
Δ_z = \(\left|\begin{array}{ccc}
1 & 2 & 11 \\
2 & 4 & 21 \\
3 & 5 & 27
\end{array}\right|\)
= 1 (108 – 105) – 2 (54 – 63) + 11 (10 – 12)
= 1(3) – 2 (-9) + 11 (-2)
= 3 + 18 – 22 = -1
By Cramer’s Rule

∴ (x, y, z) = (2, 3, 1)

Question 6.
A total of Rs 8,500 was invested in three interest-earning accounts. The interest rates were 2%, 3%, and 6%, if the total simple interest for one year was Rs 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule)
Solution:
Let the amount invested at 2%, 3%, and 6% are x, y, and z respectively.
x + y + z = 8500 ………. (1)

2x + 3y + 6z = 38000 ………. (2)
x + y = z
x + y – z = 0 ……… (3)
Here Δ = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 3 & 6 \\
1 & 1 & -1
\end{array}\right|\)
= 1 (-3 – 6) -1 (-2 – 6) + 1 (2 – 3)
= 1 (-9) -1 (-8) + 1 (-1)
= -9 + 8 – 1 = -2
≠ 0
∴ We can apply Cramer’s Rule
Δ_x = \(\left|\begin{array}{ccc}
8500 & 1 & 1 \\
38000 & 3 & 6 \\
0 & 1 & -1
\end{array}\right|\)
= 8500 (-3 – 6) -1 (-38000 – 0) + 1 (38000 – 0)
= 8500 (-9) + 38000 + 38000
= -76500 + 76000
= -500
Δ_y = \(\left|\begin{array}{ccc}
1 & 8500 & 1 \\
2 & 38000 & 6 \\
1 & 0 & -1
\end{array}\right|\)
= 1(-38000 – 0) – 8500 (-2 – 6) + 1 (0 – 38000)
= -38000 + 68000-38000
= 68000 – 76000
= -8000
Δ_z = \(\left|\begin{array}{ccc}
1 & 1 & 8500 \\
2 & 3 & 38000 \\
1 & 1 & 0
\end{array}\right|\)
= (0 – 38000) -1 (0 – 38000) + 8500 (2 – 3)
= -38000 + 38000 – 8500
= -8500
By Cramer’s Rule

∴ Amount invested at 2% is Rs 250; Amount invested at 3% is Rs 4,000 and Amount invested at 6% is Rs 4,250

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 1 Applications of Matrices and Determinants Ex 1.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1

Question 1.
Find the rank of each of the following matrices
(i) \(\left(\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right)\)
Solution:
Let A = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
Order of A is 2 × 2 [∴ P(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
5 & 6 \\
7 & 8
\end{array}\right|\)
= 40 – 42
|A| = -2 ≠ 0
There is a minor of order 2, which is not zero
ρ(A) = 2

(ii) \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & -1 \\
3 & -6
\end{array}\right|\)
= -6 – (-3)
= -6 + 3 = -3
|A| ≠ 0
There is a minor of order 2, which is not zero
∴ ρ(A) = 2

(iii) \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Solution:
Let A = \(\left(\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right)\)
Order of A is 2 × 2 [∴ ρ(A) ≤ 2]
Consider the second order minor = \(\left|\begin{array}{ll}
1 & 4 \\
2 & 8
\end{array}\right|\)
= 8 – 8 = 0
Since the second are minor vanishes, ρ(A) ≠ 2
Consider a first order minor |1| ≠ 0
There is a minor of order 1, which is not zero
∴ ρ(A) = 1

(iv) \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 2
Consider the third order minor \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
3 & 1 & -5 \\
1 & 1 & 1
\end{array}\right|\)
= 2(1 + 5) – (-1) (3 + 5) + 1 (3 – 1)
= 2 (6) + 1(8) + 1(2)
= 12 + 8 + 2
= 22 ≠ 0
There is a minor of order 3, which is not zero
∴ ρ(A) = 3

(v) \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right]\)
Order of A is 3 × 3
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
-1 & 2 & -2 \\
4 & -3 & 4 \\
-2 & 4 & -4
\end{array}\right|\)
= -1(12 – 16) – 2 (-16 + 8) – 2 (16 – 6)
= -1(-4) – 2 (-8) – 2 (10)
= 4 + 16 – 20
= 0
Since the third order minor vanishes, therefore
∴ ρ(A) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-1 & 2 \\
4 & -3
\end{array}\right|\)
= 3 – 8
= -5 ≠ 0
There is a minor of order 2, which is not zero.
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3

The number of non zero rows is 2
∴ ρ(A) = 2

The order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
3 & 1 & -5 \\
1 & -2 & 1 \\
1 & 5 & -7
\end{array}\right|\)
= 3 (14 – 5) – 1 (- 7 – 1) – 5 (5 + 2)
= 3 (9) – 1 (-8) – 5 (7)
= 27 + 8 – 35
= 0
\(\left|\begin{array}{ccc}
1 & -5 & 1 \\
-2 & 1 & -5 \\
5 & -7 & 2
\end{array}\right|\)
= 1 (2 – 35) + 5 (-4 + 25) – 1 (14 – 5)
= 1 (-33) + 5(21) – 1 (9)
= -33 + 105 – 9
= 63 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(A) = 3

Order of A is 3 × 4
∴ ρ(A) ≤ 3
Consider the third order minor \(\left|\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -1 \\
-1 & 2 & 7
\end{array}\right|\)
= 1 (28 + 2)+ 2 (-14 – 1) + 3 (-4 + 4)
= 1 (30) + 2 (- 15) + 3 (0)
= 0
\(\left|\begin{array}{ccc}
-2 & 3 & 4 \\
4 & -1 & -3 \\
2 & 7 & 6
\end{array}\right|\)
= -2 (-6 + 21) – 3 (24 + 6) + 4 (28 + 2)
= -2(15) – 3 (30)+ 4 (30)
= 0
\(\left|\begin{array}{ccc}
1 & 3 & 4 \\
-2 & -1 & -3 \\
-1 & 7 & 6
\end{array}\right|\)
= 1 (-6 + 21) – 3 (-12 – 3) + 4 (-14 – 1)
= 1 (15) -3 (-15) + 4 (-15)
= 15 + 45 – 60
= 0
\(\left|\begin{array}{ccc}
1 & -2 & 4 \\
-2 & 4 & -3 \\
-1 & 2 & 6
\end{array}\right|\)
= 1 (24 + 6) + 2 (-12 – 3) + 4 (-4 + 4)
= 1 (30) + 2 (-15) + 4(0)
= 30 – 30
= 0
Since all third order minors vanishes, ρ(A) ≠ 3
Now, let us consider the second order minors.
Consider one of the second order minors
\(\left|\begin{array}{ll}
-2 & 3 \\
4 & -1
\end{array}\right|\) = 2 – 12 = -10 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(A) = 2

Question 2.
If A = \(\left(\begin{array}{ccc}
1 & 1 & -1 \\
2 & -3 & 4 \\
3 & -2 & 3
\end{array}\right)\) and B = \(\left(\begin{array}{ccc}
1 & -2 & 3 \\
-2 & 4 & -6 \\
5 & 1 & -1
\end{array}\right)\)
find the rank of A B and the rank of B A.
Solution:

To find the rank of AB
Order of AB is 3 × 3
∴ ρ(AB) ≤ 3
Consider the third order minor |AB| = \(\left|\begin{array}{ccc}
-6 & 3 & -2 \\
28 & -12 & 20 \\
22 & -11 & 18
\end{array}\right|\)
= -6(-216 + 220) -3(504 – 440) – 2(-308 + 264)
= – 6(4) – 3(64) – 2(-44)
= -24 – 192 + 88
= 128 ≠ 0
There is a minor of order 3, which is not zero.
∴ ρ(AB) = 3
To find the rank of BA
Order of BA is 3 × 3
∴ ρ(BA) ≤ 3
Consider the third order minor |BA| = \(\left|\begin{array}{ccc}
6 & 1 & 0 \\
-12 & -2 & 0 \\
4 & 4 & -4
\end{array}\right|\)
= 6(8 – 0) – 1(48 – 0) + 0(-48 + 8)
= 6(8) – 1(48) + 0(-40)
= 48 – 48 + 0
= 0
Since the third order minor vanishes, therefore P(BA) ≠ 3
Consider a second order minor \(\left|\begin{array}{ll}
-12 & -2 \\
4 & 4
\end{array}\right|\) = -48 + 8 = -40 ≠ 0
There is a minor of order 2 which is not zero
∴ ρ(BA) = 2

Question 3.
Solve the following system of equations by rank method
x + y + z = 9, 2x + 5y + 7z = 52, 2x – y – z = 0
Solution:
x + y + z = 9
2x + 5y + 7z = 52
2x – y – z = 0
The matrix equation corresponding to the given system is

Obviously the last equivalent matrix is in the echelon form. It has three non-zero rows.
P(A) = P (A, B) = 3 = Number of unknowns
The given system is consistent and has unique solution.
To find the solution, let us rewrite the above ech-elon form into the matrix form.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 3 & 5 \\
0 & 0 & 2
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
9 \\
34 \\
16
\end{array}\right]\)
x + y + z = 9 ……..(1)
3y + 5z = 34 ……… (2)
2z = 16 ……… (3)
z = \(\frac { 16 }{ 2 }\) = 8
z = 8
Substitute z = 8 in eqn (2)
3y + 5 (8) = 34
3y + 40 = 34
3y = 34 – 40
3y = -6
y = -2
Substitute y = -2 and z = 8 in eqn (1)
x = (-2) + 8 = 9
x + 6 = 9
x = 9 – 6
x = 3
∴ x = 3, y = -2, z = 8

Question 4.
Show that the equations 5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5 are consistent and solve them by rank method.
Solution:
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given systematic


Obviously, the last equivalent matrix is in the echelon form. It has two non-zero rows.
P([A, B]) = 2 ; P(A) = 2
P(A) = P ([A, B]) = 2 < Number of unknowns
The given system is consistent and has infinitely many solutions.
The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 5 & 1 \\
0 & -11 & 1 \\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
2 \\
-3 \\
0
\end{array}\right]\)
x + 5y + z = 2 ……. (1)
-11y + z = -3 ……… (2)
let z = k

Where K ε R

Question 5.
Show that the following system of equations have unique solutions:
x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6 by rank method.
Solution:
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is

The last equivalent matrix is in the echelon form [A, B] has 3 non-zero rows and [A] has 3 non-zero rows.
p([A,B]) = 3; ρ(A) = 3
ρ([A, B]) = ρ(A) = No. of unknowns
∴ The system of equations have unique solution.

Question 6.
For what values of the parameter λ, will the following equations fail to have unique solution:
3x – y + λz = 1, 2x + y + z = 2, x + 2y – λz = -1
Solution:
3x – y + λz = 1
2x + y + z = 2
x + 2y – λz = -1
The matrix equation corresponding to the given system is

If the equations fail to have unique solution.
ρ(A) ≠ ρ(A, B)
ρ(A, B) = 3
ρ(A) ≠ 3
Therefore 7 + 2λ = 0
2λ = -7 and λ = -7/2

Question 7.
The price of three commodities. X, Y, and Z are x,y, and z respectively Mr. Anand purchases 6 units of Z and sells 2 units of Y. Mr. Amar purchases a unit of Y and sells 3 units of X and 2 units of Z. Mr. Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn Rs 5,000/-, Rs 2,000/- and Rs 5,500/- respectively. Find the prices per unit of three commodities by the rank method.
Solution:
Let the equations for Mr. Anand, Mr. Amar, and Mr. Amit are
2x + 3y – 6z = 5000
3x – y + 2z = 2000
-x + 3y + z = 5500 respectively
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
-1 & 4 & -8 \\
0 & 1 & -2 \\
0 & 0 & 7
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
3000 \\
1000 \\
3500
\end{array}\right]\)
-x + 4y – 8z = 3000 …….. (1)
y – 2z = 1000 ………. (2)
7z = 3500 ……….. (3)
Eqn (3) ⇒ z = \(\frac { 3500 }{ 7 }\)
∴ z = 500
Eqn (2) ⇒ y – 2(500) = 1000
y = 1000 + 1000
∴ y = 2000
Eqn (1) ⇒ -x + 4 (2000) – 8 (500) = 3000
-x + 8000 – 4000 = 3000
-x + 4000 = 3000
-x = 3000 – 4000
-x = – 1000
∴ x = 1000
The price of three commodities, x, y and z are Rs. 1000; Rs. 2000 and Rs. 500 respectively.

Question 8.
An amount of Rs 5,000/- is to be deposited in three different bonds bearing 6%, 7%, and 8% per year respectively. Total annual income is Rs 358/-, If the income from the first two investments is Rs 70/- more than the income from the third, then find the amount of investment in each bond by the rank method.
Solution:
Let the three different bonds be x, y, and z
x + y + z = 5000 …….. (1)

6x + 7y = 8z + 7000
6x + 7y – 8z = 7000 ……… (3)
The matrix equation corresponding to the given system is

∴ The given system is equivalent to the matrix equation
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 2 \\
0 & 0 & -16
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right] =\left[\begin{array}{c}
5000 \\
5800 \\
-28800
\end{array}\right]\)
x + y + z = 5000 ……… (1)
y + 2z = 5800 …….. (2)
-16z = -28800 ……… (3)
eqn (3) ⇒ z = \(\frac { -28800 }{ -16 }\)
∴ z = 1800
eqn (2) ⇒ y + 2(1800) = 5800
y + 3600 = 5800
y = 5800 – 3600
∴ y = 2200
eqn (1) ⇒ x + 2200 + 1800 = 5000
x + 4000 = 5000
x = 5000 – 4000
∴ x = 1000
The amount of investment in each bond is Rs 1000, Rs 2200 and Rs 1800.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 13 Organic Nitrogen Compounds Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 13 Organic Nitrogen Compounds

12th Chemistry Guide Organic Nitrogen Compounds Text Book Back Questions and Answers

Part – I Text Book Evaluation

I. Choose the Correct Answer

Question 1.
Which of the following reagent can be used to convert nitrobenzene to aniline
a) Sn/HCl
b) ZnHg/NaOH
c) LiAlH₄
d) All of these
Answer:
a) Sn/HCl

Question 2.
The method by which aniline cannot be prepared is
a) degradation of benzamide with Br₂/NaOH
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.
c) Hydrolysis of phenylcyanide with acidic solution.
d) reduction of nitrobenzene by Sn/HCl
Answer:
b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Question 3.
Which one of the following will not undergo Hofmann bromamide reaction
a) CH₃CONHCH₃
b) CH₃CH₂CONH₂
c) CH₃CONH₂
d) C₆H₅CONH₂
Answer:
a) CH₃CONHCH₃
Solution: Only primary amides undergo Hofmann bromamide reaction

Question 4.
Assertion: Acetamide on reaction with KOH and bromine gives acetic acid
Reason: Bromine catalyses hydrolysis of acetamide.
a) if both assertion and reason are true and reason is the correct explanation of assertion
b) if both assertion and reason are true but reason is not the correct explanation of assertion.
c) assertion is true but reason is false
d) both assertion and reason are false
Answer:
d) both assertion and reason are false.

Question 5.

a) bromomethane
b) α -Bromo sodium acetate
c) methanamine
d) acetamide
Answer:
c) methanamine
Solution:

Question 6.
Which one of the following nitro compounds does not react with nitrous acid?
a) CH₃-CH₂-CH₂-N0₂
b) (CH₃)₂ CH-CH₂ NO₂
c) (CH₃)₃CNO₂

Answer:
c) (CH₃)₃CNO₂
Solution:
3⁰ Nitroalkane

Question 7.

a) Friedel – Crafts reaction
b) HVZ reaction
c) Schotten – Baumann reaction
d) None of these
Answer:
c) Schotten – Baurnann reaction

Question 8.
The product formed by the reaction of an aldehyde with a primary amine (NEET)
a) carboxylic acid
b) aromatic acid
c) Schiff’s base
d) ketone
Answer:
c) Schiff’s base

Question 9.
Which of the following reaction is not correct

d) None of these
Answer:

Solution:
p – nitrosation takes places, the product is im7

Question 10.
When aniline reacts with acetic anhydride the product formed is
a) o – amirìoacetophenone
b) m – aminoacetophenone
c) p – aminoacetophenone
d) acetanilide
Answer:
d) acetanilide
Solution:

Question 11.
The order of basic strength for methyl substituted amines in aqueous solution is
a) N(CH₃)₃> N(CH₃)₂ H > N(CH₃)H₂> NH₃
b) N(CH₃)H,> N(CH₃)₂H > N(CH₃)₃ > NH₃
c) NH₃> N(CH₃)H₂> N(CH₃)₂H > N(CH₃)₃
d) N(CH₃)₂ H > N(CH₃)H₂> N(CH₃)₃ > NH₃
Answer:
d) N(CH₃)₂ H > N(CH₃)H₂> N(CH₃)₃ > NH₃

Question 12.

a) H₃PO₂ and H₂O
b) H⁺/H₂O
c) HgSO₄ / H₂SO₄
d) Cu₂Cl₂
Answer:
a) H₃PO₂ and H₂0

Question 13.

a) C₆H₅-OH
b) C₆H₅-CH₂OH
c) C₆H₅-CHO
d) C₆H₅NH₂
Answer:
a) C₆H₅-OH
Solution:

Question 14.
Nitrobenzene on reaction with Con HNO₃ / H₂SO₄ at 80- 100^oC forms which one of the following products?
a) 1, 4 – dinitrobenzene
b) 2, 4, 6 – tirnitrobenzene
c) 1, 2 – dinitrobenzene
d) 1, 3 – dinitrobenzene
Answer:
d) 1, 3 -dinitrobenzene
Solution:

Question 15.
C₅H₁₃N reacts with HNO₂ to give an optically active compound – The compound is
a) pentan-1 -amine
b) pentan-2-amine
c) N, N – dimethyipropan -2-amine
d) N-methylbutan-2-amine
Answer:
d) N-methylbutan-2-amine

Question 16.
Secondary nitro alkanes react with nitrous acid to form
a) red solution
b) blue solution
c) green solution
d) yellow solution
Answer:
b) blue solution

Question 17.
Which of the following amines does not undergo acetylation?
a) t-butylamine
b) ethylamine
c) diethylamine
d) triethylamine
Answer:
d) triethylamine (3^oamine)

Question 18.
Which one of the following is most basic?
a) 2, 4 – dichloroaniline
b) 2, 4 – dimethylaniline
c) 2, 4 – dinitroaniline
d) 2, 4 – dibromoaniline
Answer:
b) 2,4-dimethylaniline
Solution: CH₃ is a+I group, all other – I group +T group increase the electron density on NH₂ and hence increase the basis nature.

Question 19.

a) Ethanol, hydroxylamine hydrochloride
b) Ethanol, ammonium hydroxide
c) Ethanol, NH₂OH
d) C₃H₅NH₂, H₂O
Answer:
a) Ethanol, hydroxylamine hydrochloride

Question 20.
UPAC name for the amine

a) 3 – Bimethy lamino – 3 – methyl pentane
b) 3(N,N – Triethyl) – 3 – amino pentane
c) 3-N,N – trimethyl pentanamine
d) 3 – (N,N – Dimethyl amino) – 3- methyl pentane
Answer:
d) 3 – (N,N – Dimethyl amino) -3- methyl pentane

Question 21.


Answer:
b)

Question 22.
Ammonium salt of benzoic acid is heated strongly with P₂O₅ and the product so formed is reduced and then treated with NaNO₂/ HCl at low temperature. The final compound formed is
a) Benzene diazonium chloride
b) BenzYl alcohol
c) Phenol
d) Nitrosobenzene
Answer:
b) Benzyl alcohol
Solution:

Question 23.
Identify X in the sequence given below.


Answer:

Question 24.
Among the following, the reaction that proceeds through an electrophilic substitution, is:

Answer:

Explanation:
a) Nucleophilic substitution
b) Electrophilic substitution
c) Addition Reaction
d) Nucleophilic substitution

Question 25.
The major product of the following reaction



Answer:

Solution:

II. Short Answer Questions

Question 1.
Write down the possible isomers of the C₄H₉NO₂ and give their IUPAC names
Answer:

Question 2.
There are two isomers with the formula CH₃ NO₂. How will you distinguish between them?
Answer:

• Primary and secondary nitroalkanes with a-H atom exhibit tautomerism.
• Tertiary amines do not exhibit tautomerism due to CH =N the absence of a-H atom.
• Nitromethane exists in two tautomeric forms namely \itro form Aci form nitroform and aciform.

Question 3.
What happens when
Answer:
i) 2 – Nitropropane boiled with HCl
ii). Nitrobenzene electrolytic reduction in strongly acidic medium.
iii). Oxidation of tert – butylamine with KMnO₄
iv). Oxidation of acetone oxime with trifluoromethoxy acetic acid.

Question 4.
How will you convert nitrobenzene into
Answer:

• 1, 3, 5 – trinitrobenzene
• o and p – nitrophenol
• m – nitro aniline
• azoxvbenzcne
• hvdrazohenzene
• N – phenvl hyd roxylamine
• aniline

Question 5.
Identify compounds A, B, and C in the following sequence of reactions.
Answer:

Question 6.
Write short notes on the following
Answer:

• Hofmann’s bromide reaction
• Ammonolysis
• Gabriel phthalimide synthesis
• Schotten – Baurnann reaction
• Carhvlamine reaction
• Mustard oil reaction
• Coupling reaction
• Diazotisation
• Gorenberg reaction

I Hofmann’s bromamide reaction:

In Hofnianns degradation acid amide is converted into an amine with one carbon less by Br₂/ KOH.

ii. Ammonolysis:

In Hoffmann’s ammonolysis alkyl halides are heated with alcoholic ammonia in a sealed tube, mixtures of 1^o, 2^o and 3^o amines and quarternary ammonium salts are obtained.

iii. Gabriel phthalimide synthesis

Phthalimide on treatment with alcoholic KOH forms potassium phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine. Aniline can not be prepared by this method.

iv. Schotten – Baumann reaction:

Benzoylation of amines to give N-alkyl benzamide in presence of NaOH is known as Schotten Baumann reaction.

v. Carbylamine reaction

Primary’ amines react with chloroform and alcoholic KOH to form isocyanides called carbylamines with unpleasant smell. This is used to identify primary amines.

vi. Mustard oil reaction
When primary amines are treated with carbon disulphide, N-alkyl dithiocarbamic acid is formed which on treatment with HgCl₂ gives an alkyl isothiocyanate.

vii. Coupling reaction:

Benzenediazonium chloride reacts with electron-rich aromatic compounds like phenol, undergoing
electrophilic substitution at para position.

viii. Diazotisation:

Aniline reacts with nitrous acid at 273-278K to form benzene diazonium chloride.

ix. Gomberg reaction:

Benienediazonium chloride reacts with benzene in presence of NaOH to give biphenyl.

Question 7.
How will you distinguish between Primary, Secondary, and tertiary aliphatic amines
Answer:

Question 8.
Account for the following
Answer:

• Aniline does not undergo Friedel – Crafts reaction
• Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
• pK_b of aniline is more than that of methylamine
• Gabriel phthalimide synthesis is preferred for synthesising_primary amines
• Ethvlamine is soluble in water whereas aniline is not
• Amines are more basic than amides
• Although amino group is o – and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m – nitroaniline.

i) Aniline does not undergo Friedel – crafts reaction, because Aniline is basic in nature
Aniline donates its lone pair to the lewis acid Alcl₃ to form an adduct which inhibits further electrophilic substitution reaction

ii) The stability of diazonium salts of aromatic amines is due to the dispersal of the positive charge over the benzene ring.

iii) In aniline the NH
group is directly attached to the benzene ring.
The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
Hence it is less available for protonation.
So aniline is less basic than methylamine and pK_b of aniline is more than that of methylamine.

iv) Gabriel phthalimide synthesis results in the formation of primary amines only. Secondary and tertiary amines are not formed in this synthesis. Thus a pure primary amine can be obtained.
Therefore, Gabriel phthalimide synthesis is preferred for synthesizing primary amines.

v) Ethvlamine forms intermolecular hydrogen bonds with water. Hence it is soluble in water.
But aniline does not form hydrogen bond with water to a very large extent due to the presence of a large hydrophobic C₆H₅ – group. Hence aniline is insoluble in water

vi) In amides the carbonyl group (C = 0) is highly electronegative and draws the electrons towards it.
This makes the lone pair of amide nitrogen less available to accept a proton.
Hence amides are less basic than amines.
But in amines the alkyl groups being electron releasing groups. the electron pair on amine nitrogen is readily available to accept a proton.

vii) Hence amines are more basic than amides.
In strong acid medium aniline is protonated to form anilinium ion which is rn-directing.
Hence a substantial amount of rn-nitroaniline is formed during nitration of aniline.

Question 9.
Arrange the following
Answer:
i) In increasing order of solubility in water C₆H₅NH₂,(C₂H₅)₂NH,C₂H₅NH₂
ii) In increasing order of basic strength
a) aniline, p – toludine and p – nitroaniline
b) C₆H₅NH₂ C₆H₅NHCH₃, C₆H₅NH₂, p-Cl-C₆H₄-NH₂
iii) In decreasing order of basic strength in gas phase
(C₂H₅)NH₂, (C₂H₅)NH, (C₂H₅)₃N and NH₃
iv) In increasing order of boiling point C₆H₅OH, (CH₃)₂NH,C₂H₅NH₂
v) In decreasing order of the pK_b values C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and CH₃NH₂
vi) Increasing order of basic strength C₆H₅NH₂, C₆H₅N(CH₃) H₅3)₂, (C₂H₅)₂ NH and CH₃ NH₂
vii) In decreasing order of basic strength

i) Increasing order of solubility in water C₆H₅NH₂ < (C₂H₅) NH < C₂H₅ NH₂
Hint
Aromatic amine
aliphatic amine
Solubility α \(\frac{1}{\text { Molarmass }}\)
ii) Increasing order of basic strength
a) p-nitro aniline
aniline
p-toludine
b) p – Cl- C₆H₄– NH₂ < C₆H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂
Hint > p-nitrogroup (-I) electron with drawing, hence decreases the basic strength.
Hint > p-methyl group (+I) electron releasing, hence increases the basic strength.
iii) Decreasing order of basic strength in gas phase
(C₂H₅)₃N > (C₂H₅)₂ NH > C₂H₅ NH₂> NH₃
Hint > In the gas-phase there is no solvation effect. As a result the basic strength mainly depends on the + I effect. Higher the number of alkyl groups higher will be the +1 effect, stronger is the base.
iv) Increasing order of boiling point.
(CH₃)₂NH <C₂H₅NH₂ <C₆H₅OH
Hint > Amines generally have lower boiling point than alcohols of comparable molar mass. Since amines have weaker H-bonds.
Hint
Secondary amines
primary amines
v) Decreasing order of pK values
C₆H₅NHCH₃> C₂H₅NH₂> CH₃ NH₂> (C₂H₅)₂ NH
Hint > Basic character increases. pKb value decreases. Higher the basic nature, lower will be the pkb’ value.
vi) Increasing order of basic strength
C₆H₅NH₂ < C₆H₅-N(CH₃)₂ < CH₃ NH₂ < (C₂H₅)₂NH
vii) Decreasing order of basic strength

Question 10.
How will you prepare propan – 1 – amine from
Answer:
i) butane nitrile
ii) propanamide
iii) 1-nitropropane

Question 11.

Answer:

Question 12.
How will you convert diethylamine into
Answer:
i) N,N – dimethylacetamide
ii) N – nitrosódiethylamine

Question 13.
Identify A, B and C
Answer:

Question 14.
Identify A,B,C and D
Answer:

Question 15.
Complete the following reaction

Answer:

Question 16.
Predict A,B, C and D for the following reaction

Answer:

Question 17.
A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO₂ . (B) on heating with liquid ammonia followed by treating with Br₂ / KOH gives (c) which on treating with NaNO₂ and HC1 at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.
Answer:
D is a monobasic and with molecular mass 74.
M.F of D is CnH_2n+1COOH
MolarMassis 12 × n + 2n × 1 + 1 × 1 + 12 + 2 × 16 + 1 × 1 = 74
14n + 46 = 74
14n=28
n=2

Question 18.
Identify A to E in the following frequency of reactions

Answer:

III. Evaluate Yourself

Question 1.
Write all possible isomers for the following compounds.
Answer:
i) C ₂H ₅ – NO ₂
ii) C ₃H ₇ – NO ₂

Question 2.
Find out the product of the following reactions.
Answer:

Question 3.
Predict the major product that would be obtained on nitration of the following compounds?
Answer:

Question 4.
Draw the structure of the following compounds
i. Neopentylaniine
ii. Tert – butylamine
iii. α- amino propionaldehyde
iv. Tribenzylamine
v. N – ethyl – N – methyihexan – 3- amine
Answer:

Question 5.
Give the correct IUPAC names for the following amines
Answer:

12th Chemistry Guide Organic Nitrogen Compounds Additional Questions and Answers

Part – II – Additional Questions

I. Choose the correct answer

Question 1.
Nitro ethane and ethyl nitrite are
a) chain isomers
b) position isomers
c) functional isomers
d) tautomers
Answer:
c) functional isomers

Question 2.
1-nitrobutane and 2-nitrobutane are
a) chain isomers
b) Position isomers
c) functional isomers
d) tautomers
Answer:
b) Position isomers

Question 3.
1-nitrobutane and 2- methyl-1-nitro propane are
a) chain isomers
b) Position isomers
c) funtional isomers
d) tautomers
Answer:
a) chain isomers

Question 4.

a) chain isomers
b) position isomers
c) functional isomers
d) tautorners
Answer:
d) tautomers

Question 5.
Which among the following is a tertiai nitroalkane?
a) 2-nitro propane
b) 2-methyl-.1—nitro propane
C) 2-methyl – 2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl – 2-nitropropan

Question 6.
Which among the following does not sho tautomerism?
a) 2-nitro propane
b) 2-methyl-1-nitropropane
c) 2-methyl-2-nitropropane
d) nitro ethane
Answer:
c) 2-methyl-2-nitropropane
Hint : Tertiary Nitroalkanes do not exhibit tautomerism due to the absence of Oc-H atom.

Question 7.
Which nitro alkane does not dissolve in NaOH?
a) 2-nitro propane
b) 2-methyl – 1 – nitro propane
c) 2 – methvl-2-nitropropane
d) nitroethane
Answer:
c) 2 – methyl- 2 – nitropropane
Hint : Tertiary mitroalkanes do not react with NaOH due to the absence of ∝-H atom.

Question 8.
Which among the following does not react with Nitrous acid.
a) 2 – nitro propane
h) 2-methyl-1-nitropropane
c) 2 – methyl – 2 – nitro propane
d) nitro ethane
Answer:
c) 2 – methyl – 2 – nitro propane
Hint : Tertiary mitroalkanes do not react with nitrous acid due to the absence of ∝- H atom.

Question 9.
The correct order of acidic nature of nitro alkanes is
a) nitro propane> nitroethane > nitro methane
b) nitro propane c) nitro methane < nitro ethane < nitro propane d) nitro ethane> nitro methane> nitro propane
Answer:
b) nitro propane < nitroethane < nitromethane
Hint: Length of alkyl group attached to the ∝- carbon increases, acidic nature decreases

Question 10.

The above reaction follows the mechanism
a) E₁
b) E₂
c) S_N¹
d) S_N²
Answer:
d) S_N²

Question 11.
Acetaldoxime is converted into nitroethane by reaction with
a) HNO₂
b) CF₃COOOH
c) KMnO₄
d) HNO₃/H₂SO₄
Answer:
b) CF₃COOOH

Question 12.
Oil of mirbane is
a) nitro ethane
b) nitro propane
c) nitro benzene
d) nitro aniline
Answer:
c) nitrobenzene

Question 13.
Direct nitration of nitro benzene gives
a) o – dinitro benzene
b) m – dinitro benzene
c) p – dinitro benzene
d) All the above
Answer:
b) m – dinitro benzerte

Question 14.
P – diamino benzene can be converted to p – dinitro benzene by
a) Caros acid
b) persuiphuric acid
c) peroxy trifluoro acetic acid
d) all the above
Answer:
d) all the above

Question 15.
Nef carbonyl synthesis given by
a) C₆H₅CHO

c) C₆H₅CH₂NO₂
d) All of these
Answer:
a) C₆H₅CHO

Question 16.
In electrophilic substititution reaction of nitro benzene, the – NO₂ group acts as
a) activating group and m – directing
b) deactivating group and rn-directing
c) activating group and o,p – directing
d) deactivating group and o,p -directing
Answer:
b) deactivating group and m-directing

Question 17.
Identify X in the following sequence of reaction

a) (CH₃)₂CHCOCl
b) CH₃CH₂CH(OH)CH₂Cl
c) CH₃CH₂CH₂COCl
d) ClCH₂CH₂CH₂CHO
Answer:
c) CH₃CH₂CH₂COCl

Question 18.
The reducing agent used to reduce alkylcyanides to primary amines is
a)H₂/Ni
b)LiAlH₄
c) Na/C₂H₅OH
d) all the above
Answer:
d) all the above

Question 19.
The reducing agent used in Mendius reaction is
a) H₂/Ni
b) LiAlH₄
c) Na/C₂H₅OH
d) NaBH₄
Answer:
c) Na/C₂H₅OH

Question 20.
Reduction of alkyl iso cyanides give
a) primary amines
b) secondary amines
c) tertiary amines
d) none of the above
Answer:
b)secondary amines

Question 21.

a) CH₃NH₂, CH₃CH₂NH₂
b) CH₃OH, CH₂CH₂OH
c) CH₃CH₂NH₂, CH₃NH₂
d) CH₃CN, CH₃CH₂CN
Answer:
c) CH₃CH₂NH₂, CH₃NH₂

Question 22.
Gabriel phthalimide synthesis is used for the preparation of
a) aliphatic primary amines
b) aromatic primary amines
c) both (a) & (b)
d) none of the above
Answer:
a)aliphatic primary amines

Question 23.
Which among the following can not be prepared by Gabriel phthalimide synthesis?
a) Methanamine
b) ethanamine
c) benzenamine
d) propananiine
Answer:
c)benzenamine

Question 24.
Two molecules of propannitrile in the presence of Na/Ether to form 3-imino- 2-methylpentanenitrile. This reaction is known as
a) Baltz – schiemann reaction
b) Thorpe nitrile condensation
c) Gomberg reaction
d) hotten – Baurnann reaction
Answer:
b) Thorpe nitrile condensation

Question 25.
In Hoffmann’s ammonolysis, if excess alkyl halide is taken, the final product is
a) primary amine
b) secondary amine
c) tertiary amine
d) quaternary ammonium salt
Answer:
d) quaternary ammonium salt

Question 26.
Hoffmann’s ammonolysis reaction is an example of
a) nucleophilic addition
b) nucleophilic substitution
c) electrophilic addition
d) electrophilic substitution
Answer:
b) nucleophilic substitution

Question 27.
In Hoffmann’s ammonolysis the order of reactivity of alkyihalides with amines is
a) RI > RBr> RCl
b) RI < RBr < RCl c) RI > RBr < RCl
d) RI < RBr> RCl
Answer:
a)RI>RBr>RCl

Question 28.

A and B are respectively
a) CH₃Na, CH₃CN
b) CH₃NH₂,CH₃N₃
c) CH₃N₃ CH₃NH₂
d) CH₃NH₂,CH₃CN
Answer:
c) CH₃N₃ CH₃NH₂

Question 29.
The correct ascending order of basic strength of alkylamines in aqueous solution is.
a) R₂NH > RNH₂ > R₃N > NH₃
b) NH₃ > R₃N > RNH₂ > R₂NH
c) NH₃ < R₃N < RNH₂ < R₂NH
d) R₂NH < RNH₂ < R₃N < NH₃
Answer:
c) NH₃ < R₃N < RNH₂ < R₂NH

Question 30.
Among the three types of amines, secondary amine is more basic due to
a) +1 effect
b) steric effect
c) hydration effect
d) all the above
Answer:
d) all the above

Question 31.
In case of sustituted aniline, the groups which increase the basic strength are
(i) -CH₃
(ii) -NO₂
(iii) -OCH₃
(iv) -Cl
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b)(i) & (iii)

Question 32.
The correct order of basic strength of amines is
a) C₆ H₅ – NH₂ > C₆ H₅ CH₂ NH₂> NH₃ > C₆ H₅ CH₂ NHCH₃ > C₂ H₅ NH₂

b) C₆ H₅ CH₂ NH₂ > C₆ H₅ CH₂ NHCH₃ > NH₃ > C₆ H₅ NH₂ > C₂H₅NH₂

c) C₂ H₅ NH₂ > C₆H₅ CH₂ NH₂ > NH₃ > C₆ H₅ CH₂ NHCH₃ > C₆ H₅ NH₂

d) C₆ H₅CH₂ NH CH₃ > C₆ H₅ CH₂ NH₂ > NH₃ > C₂ H₅ NH₂ > C₆ H₅ NH₂
Answer:
c) C₂ H₅ NH₂ > C₆H₅ CH₂ NH₂ > NH₃ > C₆ H₅ CH₂ NHCH₃ > C₆ H₅ NH₂
Hint : Alkylamine > Aralkvlamine > Ammonia > N – Aralkylamine > Arylamine

Question 33.
Schotten Baumann reaction is an example of
a) nucleophilic substitution
b) nucleophilic addition
c) electrophilic substitution
d) electropholic addition
Answer:
a)nucleophilic substitution

Question 34.
The product ‘D’ of the reaction

a) CH₃CH₂NH₂
b) CH₃CN
c) HCONH₂
d) CH₃CONH₂
Answer:
d) CH₃CONH₂

Question 35.
The test used to identify a primary amine is
a) iodoform test
b) silver mirror test
c) libermann’s nitroso test
d) carbylamine test
Answer:
d) carbylamine test

Question 36.
The test used to identify a secondary amine is
a) iodo form test
b) silver mirror test
c) Libermann’s nitroso test
d) carbylamine test
Answer:
c) Libermann’s nitroso test

Question 37.
When ethylamine is treated with chloroform and alcoholic KOH an unpleasant smell is evolved due to the formation of
a) ethylcyanide
b) ethylisocyanide
c) ethylisocyanate
d) ethylcyanate
Answer:
b) ethylisocyanide

Question 38.
C₆H₅N + 2Cl^– + H₃PO₂ + H₂O → C₆H₆ + H₃PO₃ + HCl + N₂ This reaction proceeds through
a) SN¹ mechanism
b) SN² mechanism
c) free radical mechanism
d) elimination reaction
Answer:
c) free radical mechanism

Question 39.
Benzene diazonium chloride is converted into chloro benzene by
(i) Sandmeyer reaction
(ii) Gatter mann reaction
(iii) Gomberg reaction
a) (i) &(ii)
b) (i.) & (iii)
c) (ii) & (iii)
d) (i) (ii) & (iii)
Answer:
a)(i) &(ii)

Question 40.


Answer:

Question 41.
Hydrogen cyanide and hydrogen isocyanide are
a) chain isomers
b) position isomers
c) tautomers
d) Geometrical isomers
Answer:
c) tautomers

Question 42.
The IUPAC name of C₆H₅CN is
a) phenyl cyanide
b) benzene carbonitrile
c) benzonitrile
d) benzene cyanide
Answer:
b) benzene carbonitrile

Question 43.

a) CH₃CN
b) CH₃NH₂
c) CH₃CONH₂
d) CH₃CH₂CONH₂
Answer:
d) CH₃CH₂CONH₂

Question 44.
Methyl magnesium chloride is converted into ethanenitrile by treating with
a) cyanogen
b) cyanogen chloride
c) methyl cyanide
d) hydrogen cyanide
Answer:
b) cyanogen chloride

Question 45.
Which one of the following compound is a strong base?

Answer:

Question 46.
Which among the following will not undergo diazotisation?
a) m- toluidine
b) aniline
c) p – amino phenol
d) benzylamine
Answer:
d) benzylamine
Hint: Aromatic amines in which – NH₂ group is directly attached to the benzene ring undergo diazotisation

Question 47.
An organic compound (A) C₃H₉N when treated with nitrous acid, gave an alcohol (B) and N₂ gas. (A) on warming with CHCl₃ and Caustic potash gave (C) which on reduction gave isopropyl methylamine. Organic compound (A) is
a) CH₃CH₂NH CH₃
b) CH₃ CH₂ CH₂ NH₂

Answer:

Question 48.

a) Ethanamide, Ethanamine, Ethanol
b) Ethanarnine, Ethanol, Ethanoic acid
c) Ethanol, Ethanal, Ethanoic acid
d) Ethanamine, Ethanal, Ethanoic acid
Answer:
b) Ethanamine, Ethanol, Ethanoic acid
Hints:

Question 49.
Which of the following nitro compounds when reacted with nitrous acid followed by treatment with alkali produces blue colour?
a) 2-methyl-2-nitropropane
b) 2-methyl-1-nitropropane
c) 2-nitropropane
d) nitrobezene
Answer:
c) 2-nitropropane
Hunt: Nitro group attached to a secondary carbon atom will give pseudonitrol with HNO₂ which on further reaction with alkali gives blue colour.

II. Assertion and reason

Question 1.
Assertion (A) : Aniline when exposed to air becomes coloured
Reason (R) : Aniline when exposed to air undergoes oxidation
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both A and R are correct, R explians A

Question 2.
Assertion (A) : Aniline is more basic than ammonia
Reason (R) : The lone pair of electron on nitrogen atom in aniline gets delocalised over the benzene ring.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
d) A is wrong but R is correct.
Correct Assertion A: Aniline is less basic than ammonia.

Question 3.
Assertion (A) : Nitroalkanes are more acidic than aldehvdes, ketones and cyanides.
Reason (R) : a-H atom of 10 and 20 nitroalkanes, show acidic character because of the electron releasing effect of NO₂ group.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
c) A is correct but R is wrong.
Correct Reason (R) : a- H atom of 1^o and 2^o nitroalkanes show acidic character because of the electron withdrawing effect of NO₂ group.

Question 4.
Assertion (A) : Nitro benzene does not undergo Friedel crafts reactions:
Reason (R) : NO₂ group present in the benzene ring is strongly deactivating in nature.
a) Both A and R are correct, R explains A
b) Both A and R are correct, R does not explain A
c) A is correct but R is wrong
d) A is wrong but R is correct
Answer:
a) Both(A) and (R) are correct, R explains A.

III. Pick out the correct statement

Question 1.
(i) In amines the nitrogen atom is Sp₃ hybridised.
(ii) In amines, the fourth Sp₃ hybridised orbital contains a lone pair of electron.
(iii) Due to the presence of lone pair of electron C- N-C bond angle is more than the normal tetrehedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1090 which is greater than tetrahedral angle,
(a) (i) & (ii)
(b) (ii)&(iii)
(c) (iii)&(iv)
(d) (I&(iv)
Answer:
(a)(i) & (ii)
Correct statement:
(iii) Due to the presence of lonepair of electron C-N-C bond angle is less than the normal tetrahedral bond angle.
(iv) The C-N-C bond angle of trimethylamine is 1080 which is lower than tetrahedral angle.

Question 2.
(i) Tertiary amines form inter molecular hydrogen bond.
(ii) Amines have higher boiling points than alcohols.
(iii) Nitrogen being less electronegative than oxygen, the N-H bond in amines is less polar than O-H bond in alcohols
(iv) Lower aliphatic amines are soluble in water, since they form hydrogen bond with water,
(a) (i) & (ii)
(b) (ii) & (iii)
(c) (iii) & (iv)
(d) (i) & (iv)
Answer:
(c) (iii) & (iv)
Correct Statement:
(i) Tertiary amines do not form intermolecular hydrogen bond.
(ii) Amines have lower boiling point than alcohols.

Question 3.
(i) In Hofmann’s degradation reaction primary amines with one carbon less than the parent amides are formed.
(ii) Gabriel phthalimide synthesis is useful for the preparation of three types of amines.
(iii) Alkyl halides can be converted to primary amines by treating with sodium azide followed by reduction using LiAlH₄
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl₂ gives p -amino phenol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (i) & (iv)
Answer:
c)(i) & (iii)
Correct statement:
(ii) Gabriel phthalimide synthesis is useful for the preparation of aliphatic primary amines
(iv) When phenol reacts with ammonia in presence of anhydrous ZnCl₂ gives aniline.

Question 4.
(i) Alkyl cyanides on reduction with LiAlH₄ give secondary amines.
(ii) Alkyl cyanides on hydrolysis with alkali or dilute mineral acid give carboxylic acids.
(iii) Nitriles containing a- H atom undergo condensation with esters in presence of sodamide in ether to form ketonitriles.
(iv) Alkycyanides have lower boiling points than analogous acetylenes
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d (i) & (iv)
Answer:
b)(ii) & (iii)
Correct statement:
(i) Alkylcyanides on reduction with LiAlH₄ give primary amines.
(iv) Alkylcyanides have higher boiling points than analogous acetylenes.

IV. Pick out the incorrect statement

Question 1.
(i) Bromobenzene can be converted into nitrobenzene by reacting with KNO₂
(ii) Nitroethane is suspected to cause genetic damage and be harmful to the nervous system.
(iii) 2-methyl – 2-nitro propane reacts with HCl undergoing hydrolysis to form 2-methyl – 2 – propanol.
(iv) Ethyl nitrite on reduction with Sn / HCl gives ethanol.
a) (i) & (ii)
b) (ii) & (iii)
c) (i) & (iii)
d) (iii) & (iv)
Answer:
c) (i) & (iii)
Correct statement:
(i) Bromo benzene can not be converted into nitrobenzene by reacting with KNO₂
(iii) 2-methyl -2-nitropropane does not undergo hydrolysis with HCl to form 2-methyl – 2 – propanol

Question 2.
(i) Nitroalkanes exhibit functional isomerism with alkyl nitrites.
(ii) Tertiary nitroalkanes exhibit tautomerism due to the absence of a-H atom.
(iii) The electrical conductivity of aciform of nitro alkanes is high.
(iv) When the number of alkyl group attached to the a- carbon of nitroalkane increases, its acidity increases.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statemet:
(ii) Tertiary nitroalkanes do not exhibit tautomerism due to the absence of a- H atom.
(iv) When the number of alkyl group attached to the a- carbon of nitro alkane increases its acidity decreases.

Question 3.
(i) Benzoylation of aniline is known as Schotten. Baumann reaction
(ii) The reaction of ethylamine with nitrous acid is known as diazotisation.
(iii) The reaction of primary amine with nitrous acid is known as Libermann’s nitroso test.
(iv) The reaction of primary amine with CS₂ and HgCl₂ is known as Mustard oil reaction.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statement:
(ii) The reaction of aniline with nitrous acid is known as diazotisation.
(iii) The reaction of secondary amine with nitrous acid is known as Libermann’s nitroso test.

Question 4.
(i) Diazo compounds obtained from the coupling reactions of diazonium salts are coloured and are used as dyes.
(ii) Aryl fluorides and iodides can not be prepared by direct halogenation but can be prepared by using benzene diozonium chloride.
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu/HCl and Cu₂Cl₂ / HCZ respectively .
(iv) Benzene diazonium chloride when added with boiling water gives benzene.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statement:
(iii) In Sandmeyer reaction and Gattermann reaction the catalyst used are Cu₂Cl₂/ HCl and Cu / HCl respectively
(iv) Benzene diazonium chloride when added with boiling water gives phenol.

V. Match the following

Question 1.

Answers:
1. – Phenylhydroxylamine
2. – Azobenzene
3. – Aniline
4. – Hydrazo benzene
5. – Nitrosobenzene

Question 2.


Answer:
1. – Prop – 2-en-1-amine
2. – N – methylpropan – 2 – amine
3. – Hexan -1, 6 – diamine
4. – Phenylmethanamine
5. – N, N – dimethylben zenamine

Question 3.

Answer:
1. – N-phenylbenzamide
2. – Benzene diazonium chloride
3. – Methyl isocyanide
4. – Methylisothiocyanate
5. – p-hydroxyazobenzene

Question 4.

Answer:
1. – insecticide
2. – diuretic
3. – TNT
4. – fuel additive
5. – Solvent in perfume industry

VI. Two Mark Questions

Question 1.
Write a note on acidic nature of nitroalkanes.
Answer:

• α-H atom of 1⁰and 2⁰nitroalkanes show acidic character.
• This is due to the electron withdrawing effect of – NO₂group.
• Nitroalkanes are more acidic than aldehydes, Ketones, esters and cyanides.
• Due to acidic nature nitroalkanes dissolve in NaOH to form a salt.
• Aci-nitro form is more acidic than nitro form.
• As the number of alkyl group attached to a- Carbon increases, acidity decreases due to + I effect of alkyl groups.

Question 2.
How is the oil of mirbane prepared?
Answer:
When benzene is heated at 330K with the nitrating mixture (Con HNO₃ + Con H₂S0₄), electrophilic substitution takes place to form nitrobenzene (oil of mirbane)

Question 3.
How is P – dinitrobenzene prepared?
Answer:

• Direct nitration is nitrobenzene gives m-dinitro benzene.
• The following indirect method is adopted to prepare p-dinitrobenzene.
  

Question 4.
Convert p-diamino benzene into p-dinitro benzene.
Answer:
Amino group can be directly converted into nitro group using Caro’s acid (H₂SO₅) or persulphuric acid (H₂S₂O₈) or peroxytrifluro acetic acid (CF₃COOOH) as oxidising agent

Question 5.
Write the various reduction stages of nitromethane.
Answer:

Question 6.
Write about the reduction of nitro methane in acid medium and neutral medium.
Answer:

Question 7.
What happpens when ethylnitrite is reduced by Sn / Hcl?
Answer:

• Ethyl nitrite on reduction with Sn / HCl gives ethanol.

Question 8.
What happens when ethyl nitrite undergoes hydrolysis with acid or base?
Answer:

• Ethyl nitrite undergoes hydrolysis with acid or base to give ethanol.

Question 9.
What is Nef Carbonyl Synthesis?
Answer:

Question 10.
Write about the electrolytic reduction of nitro benzene?
Answer:

Question 11.
Convert m- dinitrobenzene into m – nitroaniline
Answer:
Selective reduction of rn – dinitrobenzene with ammonium suiphide gives m – nitroaniline

Question 12.
What happens when
(i) methyl cyanide and
(ii) methyl isocyanide are reduced?
Answer:
(i) Methyl cyanide on reduction gives a primary amine ethanamine.

(ii) Methylisocyanide on reduction gives a secondary amine N-methylmcthanamine

Question 13.
How can you convert acetamide into
(i) ethanamine
(ii) methanamine.
Answer:
Acetamide on reduction with LiAlH₄ gives ethanamine.

Acetamide undergoes Hofmanns degradation with Br₂/KOH to form methanamine

Question 14.
Write about Sabatier – Mailhe method of preparing three types of Amines
Answer:

• Vapour of an alcohol and ammonia are passed over alumina, W₂O₅ or Silica as 400_o Call the types of amines are formed.
• This method is known as Sabatier-Mailhe method.

Question 15.
Write a note on basic strength of aniline.
Answer:

• Aniline is less basic than alkylamines and ammonia.
• In aniline, the NH2 group is directly attached to the benzene ring.
• The lone pair of electron in nitrogen atom in aniline gets delocalised over the benzene ring.
• Hence it is less available for protonation.
• In case of subtituted aniline, election releasing groups like – CH₃, -OCH₃, -NH₂ increase the basic strength.
• Electron with drawing groups like -NO₂, -X, -COOH decrease the basic strength.

Question 16.
How is sulphanilic acid prepared?
Answer:

• Aniline reacts with conc H₂SO₄to form anilinium hydrogen sulphate.
• This on heating with H₂SO₄ at 453-473K gives p-aminobenzene suiphonic acid known as
  sulphanilic acid

Question 17.
How will you convert aliphatic carboxylic acid into aromatic carboxylic acid?
Answer:

• When benzene diazonium fluoroborate is heated with acetic acid, benzoic acid is obtained.
• This reaction is used to convert aliphatic carboxylic acid into aromatic carboxylic acid.

Question 18.
Write about the isomerisation of alkylisocyanides.
Answer:
When alkyl isocyanides are heated at 250_oC, they change into the more stable isomeric cyanides.

Question 19.
Write the uses of nitroalkanes.
Answer:

• They are good solvents for a large number of organic compounds including vinyl polymers, cellulose esters, synthetic rubbers, oils, fats, waxes and dyes.
• Used in organic synthesis.
  Nitromethane reacts with halogen in presence of alkali to form trihalogen derivative, (e.g.,) with chlorine it forms chloropicrin, CCl₃NO₂ which is used as soil sterilizing agent.
• Nitroethane is also used as a fuel additive and precursor to Rocket propellants.

Question 20.
How is chloropicrin prepared ?
Answer:
Nitromethane on treatment with Cl2 in presence of NaOH gives chloropicrin.

Chloropicrin (Trichloronitro methane)

VII. Three mark questions

Question 1.
How are nitro compounds classified?
Answer:
Classification of nitrocompounds

Question 2.
Convert (i) acetaldelyde into nitro ethane
(ii) acetone into 2-nitropropane
Answer:

Question 3.
Write about the hydrosysis of nitro alkanes
Answer:

• Hydrolysis can be effected using conc: HCI or conc H2S04
• Primary nitro alkanes give carboxylic acids.
• Secondary nitro alkanes give ketones.
• Tertiary nitro alkanes have no reaction.

Question 4.
Write about the classification of amines
Answer:

Question 5.
Write notes on structure of amines
Answer:

• Like ammonia, nitrogen atom of amines is trivalent.
• It carries a lone pair of electron.
• Nitrogen atom is sp³ hybridised.
• Out of four sp³ hybridised orbitals, three sp3 orbitals overlap with hydrogen or carbon of alkyl groups.
• Fourth sp³ orbital contains a lone pair of electron.
• Amines posses pyramidal geometry.
• Due to the presence of lone pair of electron the C-N-H or C-N-C bond angle in less than the normal
  tetra hedral bond angle 109.5^O
• In trimethyl amine the C-N-C bond angle is 108^O which is lower than the tetrahedral angle and higher than the H-N-H bond angle of 107^O in ammonia.
• This increase is due to the repulson between the bulky methyl groups.
  

Question 6.
Prepare aniline from
(i) chloro benzene
(ii) Phenol
(iii) nitro benzene
Answer:

Question 7.
Derive an expression for basic strength of amines.
Answer:

• In aqueous solutions the following equilibrium exists.
• It lies far to the left, hence amines are weak bases compared to NaOH.

• Basicity constant K _bis a measure of the extent to which the amine accepts the hydrogen – ion H⁺ from water.
• Larger the value of K_b,, smaller is the value of p^K_b and stronger is the base.

Question 8.
An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl₃ and alcoholic KOH gives (C) . (C) on catalytic reduction gives N-methyl aniline. Identify A, B, C and write its equation.
Answer:

Question 9.
Convert aniline into
(i) p-bromo aniline
(ii) p-nitro aniline
Answer:
(i) To get mono bromo derivatives, -NH₂ is first acylated to reduce its activity

(ii) To get para product, the -NH₂ group is protected by acetylation with acetic anhydride. Then the nitrated product is hydrolysed to form the product

VIII. Five mark questions

Question 1.
Explain structural isomerism exhibited by nitro alkanes
Answer:

Question 2.
Explain the reduction reactions of nitro benzene in various medium.
Answer:

Question 3.
An Organic compound (A) – C₇H₇NO on treatment with Br₂ and KOH gives an amine (B), which gives carbylamine test. (B) upon diazotization to give (C). (C) on coupling_with p-cresol to give
compound (D). Identify A, B, C and D with necessary reaction.
Hofmann’s degradation reaction
Compound (A is Benzamide
Compound B is Aniline (primary amine)
Diazotization reaction:

Question 4.
Write a short note an Libermann’s nitroso test ?
Answer:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine

Question 5.
An Organic compound (A) – CNCl react with methyl magnesium Bromide to give, compound B – (C₂H₃N). B – upon catalytic reduction to give compound C – (C₂H₇N). C gives carbylamine test, Identify compound A, B and C and write the reactions.
Answer:
An organic Compound A is cyanogon chloride react with methylmagnesium Bromide to give compound B – ethanenitrite – (C₂H₃N)

B – upon catalytic reduction to give compound C – Ethananxuie (C₂H₇N)

Question 6.
Explain the action of nitrous acid on three types of amine.
Answer:
Three classes of amines react differently with nitrous acid which is prepared in situ from a mixture of NaNO₂ and Hcl.
(a) Primary amines:
Ethylamine reacts with nitrous acid to form ethyldiazonium chloride which is unstable. It is converted into ethanol by liberating N₂.

• Aniline reacts with nitrous acid at low temperature (273 – 278K) to give benzene diazonium chloride.
• Benzene diazonium chloride is stable for short time and slowly decomposes even at low
  temperatures.

b) Secondary amines:
Alkyl and aryl secondary amines react with nitrous acid to give N-nitroso amine, which is insoluble in water.

c) Tertiary amines: :
Aliphatic tertiary amine reacts with nitrous acid to form trialkyl ammonium nitrite salt which is
insoluble in water.

Aromatic tertiary amine reacts with nitrous acid at 273K to give p-nitroso compound

Question 7.
Explian:
(i) Sand mayer reaction
(ii) Gaftermann reaction
(iii) Baltz – schiemann reaction
Answer:
(i) Sand mayer reaction:

(ii) Gattermann reaction:

(iii) Baltz – Schiemann Reaction:

Question 8.
Convert benzene diazonium chloride into:
(i) benzene
(ii) iodo benzene
(iii) phenol
(iv) nitrobenzene
(v) phenyl hydrazine

Question 9.
Explain coupling reactions of benzene diazonium chlorid
Answer:

• Benzene diazonium chloride reacts with electron rich aromatic compounds like phenol, aniline to form brightly coloured azo compounds.
• Coupling generally occurs at the para position.
• If para position is occupied then coupling occurs at the ortho position.
• Coupling tendency is enhanced if an electron donating group is present at the para-position to – N₂ Cl group.
• This is an electrophilic substitution.

Question 10.
Write notes on
(i) Thrope nitrile condensation
(ii) Levine and Hauser acetylation
Answer:
(i) Thrope nitrile condensation :
Self condensation of two molecules of alkylnitrile containing α-H atom in the presence of sodium to form iminonitrile.

Nitriles containing α-H atom undergo condensation with esters in the presence of sodamide in ether to form ketonitriles,
This is known as Levine – Hauser acetylation.
This reaction involves the replacement of ethoxy (OC₂H₅) group by methylnitrile (-CH₂ CN) group and is called as cyanomethylation reaction.

Question 11.
An aromatic nitro compound (A) on reduction with Sn/HCl gives compound (B) C₆H₇N which on treatment with Benzoyl chloride in the presence of pyridine to give compound (C). Compound (B) on treatment with CH₃Br to give compound (D) which further reacts with NaNO₂ /HCl to give compound (E) with yellow oil liquid. Identify (A) to (E) and write the
reactions.
Compound (A) is nitro benzene on reduction with Sn/HCl gives compound (B) is Aniline

Compound (B) is Aniline reacts with Benzoyl chloride in the presence of pyridine to give compound
(C) is N – phenyl benzamide
Schotten- Baumann reaction