Category: Important Questions

TN State Board 11th Chemistry Important Questions Chapter 1 Basic Concepts of Chemistry and Chemical Calculations

Question 1.
Give a brief account of classification of matter.
Answer:
(i) Matter can be classified as solids, liquids, and gases based on their physical state.
(ii) Matter can also be classified into mixtures and pure substances based on chemical composition.

Question 2.
Distinguish between a homogeneous and heterogeneous mixture.
Answer:

Homogeneous	Heterogeneous
A homogeneous mixture is one in which the components are uniformly mixed. Every portion of the mixture has the same compositon and proportion, (isotropic)	A heterogeneous mixture is not uniform. The properties of the mixture is not uniform throughout, (anisotropic)

Question 3.
Define the following:
(i) Element,
(ii) Compound.
Answer:
(i) An element consists of only one type of atom. An atom is the smallest electrically neutral particle, being made up of fundamental particles, namely electrons, protons, and neutrons.
(ii) Compounds are made up of molecules which contain two or more atoms of different elements. Properties of compounds are different from those of their constituent elements. The constituents of a compound is present in a fixed ratio by weight.

Question 4.
Give two examples of elementary molecules which are
(i) Monoatomic
(ii) Diatomic and
(iii) Polyatomic.
Answer:
(i) Monoatomic molecules: Copper (Cu) and Gold(Au)
(ii) Diatomic molecules: Hydrogen (H₂) and Oxygen (O₂)
(iii) Polyatomic molecules: Phosphorous (P₄) and Sulphur (S₈)

Question 5.
Mention the characteristics of a compound.
Answer:
(i) The compound is made up of two or more atoms of different elements in a definite ratio by weight.
(ii) The properties of a compound is different from those of their constituent elements.

Question 6.
Why atomic masses are called as relative atomic masses?
Answer:
The mass of an atom (10^-27 kg) is too small to be measured directly. They are measured with reference to a standard atom. Hence they are called relative atomic masses.

Question 7.
Which atom is considered to be a standard atom by IUPAC to measure atomic masses and what is its atomic mass?
Answer:
The C-12 atom is considered as standard by the IUPAC (International Union of Pure and Applied Chemistry), and it’s mass is fixed as 12 amu (or) u.

Question 8.
Define a.m.u or unified atomic mass.
Answer:
The quantity \(\frac{1}{12}\) mass of an atom of carbon -12 is known as the atomic mass unit, (a.m.u). The actual mass of one atom of carbon-12 is 1.9914 × 10^-23 g or 1.9914 × 10^-26 kg.
Thus,

1 a.m.u = \(\frac{1.9924 \times 10^{-23}}{12}\)
= 1.66 × 10^-24 g = 1.66 × 10^{-27 kg}

∴ Atomic mass of an element = \(\begin{gathered}
\text { Mass of one atom } \\
\text { of the element } \\
\hline 1 \text { a.m.u }
\end{gathered}\)

Question 9.
Define relative atomic mass.
Answer:
The relative atomic mass is defined as the ratio of the average atomic mass factor to the unified atomic mass unit.

Question 10.
Calculate the mass of the following atoms in a.m.u (unified atomic mass).
Answer:
(i) Oxygen: average mass of oxygen atom = 2.656 × 10^-23 g
(1 a.m.u= 1.6605 × 10^-27 kg (or) 1.6605 × 10^-24 gram).

(ii) Mass of an oxygen atom = 2.656 × 10^-23 g
The mass of oxygen atom in a.m.u = Average mass of oxygen atom in gram / (1.66 × 10^-23)
= \(\frac{2.656 \times 10^{-23} \mathrm{~g}}{1.6605 \times 10^{-24} \mathrm{~g}}\)
= 1.599 × 10
= 15.99 a.m.u

Question 11.
The atomic mass of oxygen is 16. What is the actual mass of an oxygen atom?
Answer:
Actual mass of an oxygen atom = The atomic mass of an element in a.m.u × 1.66 × 10^-24 g
= 16 × 1.66 × 10^-24
= 2.656 × 10^-23 g

Question 12.
The relative abundance of ₆C¹², ₆C¹³ and ₆C¹⁴ are 98.892%, 1.108% and 2 × 10^-10 respectively. Calculate the average atomic mass of carbon.
Answer:
Average atomic mass of carbon = (12 × % abundance of C¹² + 13 × % abundance of C¹³ +14 × % abundance of C¹⁴) / 100
given % abundances of C¹² = 98.892
C¹³ = 1.108
C¹⁴ = 2 × 10^-10
Since 2 × 10^-10 is so small compound to 100, it can be neglected.

Avereage atomic mass of carbon = \(\frac{12 \times 98.892+13 \times 1.108}{100}\) = 12.01 a.m.u

Question 13.
The relative abundance of ₁₀Ar³⁶, ₁₀Ar³⁸, ₁₀Ar⁴⁰ are 0.337%, 0.063% and 99.6% respectively. Calculate the average atomic weight of Argon.
Answer:
Average atomic weight of Argon = (36 × % abundance of Ar³⁶ + 38 × % abundance of Ar³⁸ + 40 × % abundance of Ar⁴⁰) / 100
given % abundances of Ar³⁶ = 0.337
Ar³⁸ = 0.063
Ar⁴⁰ = 99.6

Question 14.
Define relative molecular mass.
Answer:
Relative molecular mass is defined as the ratio of the mass of a molecule to the unified atomic mass unit.

Question 15.
Calculate the molecular masses of the following:

(i) KMn04
Answer:
Molecular mass of KMnO₄ = 1 × atomic mass of K +1 × atomic mass of Mn + 4 × atomic mass of O atomic mass of K = 39.098 ≈ 39
atomic mass of Mn = 54.938 ≈ 55
atomic mass of ‘O’ = 15.99 ≈ 16
∴ Molecular mass of KMn04. = 39 + 55 + 4 × 16 = 158 u.

(ii) NaOH
Molecular mass of NaOH = 1 × atomic mass of Na + 1 × atomic mass of O + 1 × atomic mass of H
= 1 × 23 + 1 × 16 + 1 × 1 = 23 + 16 + 1 = 40u.

(iii) H₂C₂O₄.2H₂O (crystalline oxalic acid)
Molecular mass of H₃C₂O₄.2H₂O = 2 × atomic mass of H + 2 × atomic mass of C + 4 × atomic mass of O + 2 × molecular mass of H₂O
= 2 × 1 + 2 × 12 + 4 × 16 + 36
= 2 + 24 + 64 + 36
= 126u.

(iv) C₆H₁₂O₆ (glucose)
Molecular mass of C₆H₁₂O₆ = 6 × atomic mass of C + 12 × atomic mass of H + 6 × atomic mass of O
= 6 × 12 + 12 × 1 + 6 × 16
= 72 + 12+ 96 = 180u

Question 16.
Explain the term “mole”.
Answer:
(i) Mole is the SI unit to represent a specific amount of a substance.
(ii) One mole is the amount of substance of a system which contains as many elementary particles as there are in 12g of carbon-12-isotope.
(iii) Mole represents the mass in gram of 6.022 × 10²³ entities (atom or molecule /ion).
(iv) 1 mole of any substance contains 6.023 × 10²³ entities.

Question 17.
Give the relationship between the number of mole of the substance and its gram molecular mass.
Answer:

Question 18.
Calculate the number of atoms / molecules present in the following:

(i) 10 gram of mercury
Atomic mass of Hg = 200.59
Gram atomic mass of Hg = 200.59 g
1 mol of Hg = 200g = 6.023 × 10²³ atoms
10 g = \(\frac{6.023 \times 10^{23}}{200}\) × 10
= 0.301 × 10²³
= 3.01 × 10²² atoms

(ii) 1.8 gram of water
Molecular mass of H₂O = 18
gram molecular mass of H₂O = 18 g
1 mol of H₂O = 18g of H₂O
= 6.023 × 10²³ molecules of H₂O
1.8 g of H₂O = \(\frac{6.023 \times 10^{23}}{18}\) × 1.8
= 6.023 × 10²² molecules

(iii) 100 gram of sulphur dioxide
Molecular formula of sulphur dioxide = SO₂
Molecular mass of sulphur dioxide = 64
1 mole of SO₂ = 64g of SO₂
= 6.023 × 10²³molecules
100g of SO₂ = \(\frac{6.023 \times 10^{23}}{64}\) × 100
= 9.41 × 10²³ molecules.

(iv) 1 kg of acetic acid (CH₃COOH)
Molecular mass of acetic acid = 60
1 mol of acetic acid= 60g of acetic acid
= 6.023 × 10²³ molecules
1000g of acetic acid = \(\frac{6.023 \times 10^{23}}{60}\) × 1000
= 1.0 × 10²⁵ molecules

(v) 46 gram of sodium
Atomic mass of sodium = 23 a.m.u
1 mol of Na = 23g = 6.023 × 10²³ atoms
46g = \(\frac{6.023 \times 10^{23}}{23}\) × 46
= 12.046 × 10²³ atoms.

Question 19.
Calculate the number of moles present in the following:

Number of mol = (Weight of the substance in g) / (gram molecular weight (or) gram atomic weight)

(i) 50g of CaCl₂
Weight of CaCl₂ = 50g ,
gram moleular weight = 40 + 71 = 111
∴ number of mol = \(\frac{50}{111}\) = 0.45 mol

(ii) 120g of NaOH
Weight of NaOH = 40g
gram moleular weight = 40 g
∴ number of mol = \(\frac{40}{40}\) = 1 mol

(iii) 46g of ethanol
Weight of ethanol = 46 g
gram molecular mass of ethanol (C₂H₅OH) = 46
∴ number of mol = \(\frac{46}{46}\) = 1 mol

(iv) 90 g of magnesium oxide
Weight of MgO = 90 g
gram molecular weight of MgO = 24 + 16 = 40
∴ number of mol = \(\frac{90}{40}\) = 2.25 mol

(v) 19.5 g of potassium
Weight of potassium = 19.5 g
gram atomic mass of potassium = 39
∴ number of mol = \(\frac{19.5}{39}\) = 0.5 mol

Question 20.
Define Avogadro number.
Answer:
The total number of entities (atoms, molecules, ions, electrons etc…) present in one mole of any substance is equal to 6.022 × 10²³. This number is called Avogadro number.

Question 21.
State Avogadro hypothesis.
Answer:
Avogadro hypothesis that equal volume of all gases under the same conditions of temperature and pressure contain equal number of molecules.

Question 22.
The relative molecular mass of sulphuric acid (H2S04) is 98 a.m.u. What is its molar mass?
Answer:
Molar mass of H₂SO₄ = 98g mol^-1.

Question 23.
Define molar volume.
Answer:
The volume occupied by one mole of any substance in the gaseous state at a given temperature and pressure is called molar volume. The standard temperature is 273K (0°C) and the standard pressure is 1 atm.

Question 24.
Calculate the number of moles present in 9 grams of ethane.
Answer:
Molecular formula of ethane is C₂H₆
Motecuhr weight of ethane = 2 × 12 + 6 × 1 = 30 u
Molecular mass of ethane = 30g of mol^-1
Number of mol = \(\frac{\text { Weight of substance }}{\text { Molar mass }}\)
= \(\frac{9}{30}\) = 0.3 mol

Question 25.
Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273K at 3 atm.
Answer:
Step – 1:
To find the volume O₂ at 273K at 1 atm
\(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)

P₁ = 3atm: V₁ =224 ml: T₁ = 273K
P₂ = 1 atm: V₂ = ? T₂ = 273K
\(\frac{3 \times 224}{273}=\frac{1 \times \mathrm{V}_{2}}{273}\)

V₂ = 3 × 224 = 672ml at NTP

Step-2:
22400 ml of O₂ at NTP = 1 mol of O₂
= 6.023 × 10²³ molecules
672 ml at NTP = \(\frac{6.023 \times 10^{23}}{22400}\) × 672
= 0.18 × 10²³ molecules
= 1.8 × 10²² molecules.

Question 26.
Calculate the volume occupied at STP by the following:

(i) 88 gram of CO₂
Answer:
1 mol of CO₂ = 44gof CO₂
= 22400 cm³ at STP
88g of CO₂ = 2 × 22400 cm³ = 44800 cm³

(ii) 0.5 mole of methane
Answer:
1 mol of CH₄ = 22400 cm³ at NTP
0.5 mol of CH₄ = 22400 × 0.5.
= 11200 cm³ at NTP

(iii) 3.0115 × 10²³ molecules of sulphur dioxide.
1 mol of SO₂ = 6.023 × 10²³ molecules of SO₂
= 22400 cm³ at STP
3.0115 × 10²³ molecules of SO₂ = \(\frac{22400 \times 3.0115 \times 10^{23}}{6.023 \times 10^{23}}\)
= 11200 cm³ or 11.2 L at STP

Question 27.
Define gram equivalent mass of an element.
Answer:
Gram equivalent mass of an element, compound or ion is the mass of their species that combines or displaces 1.008g hydrogen or 8g oxygen or 35.5g chlorine.

Question 28.
What is equivalence factor ‘n’? How is it useful in determining equivalent mass of a substance?
Answer:
Equivalence factor (n) = \(\frac{\text { Molar mass }\left(\mathrm{g} \mathrm{mol}^{-1}\right)}{\text { equivalent mass }\left(\mathrm{g} \mathrm{eq}^{-1}\right)}\)

The usefulness of this factor is that the equivalent masses of all the substances can be calculated whether it is an acid, base, salt, or an oxidising or reducing agent.

Question 29.
What is the equivalence factor (n) for

(i) H₂SO₄
Answer:
‘n’ factor for H₂SO₄ = \(\frac{\text { molar mass of the acid }}{\text { equivalent mass of the acid }}\)
= \(\frac{98}{49}\) = 2

(ii) Ba(OH)₂
Answer:
‘n’ factor for Ba(OH)₂ = \(=\frac{\text { molar mass of } \mathrm{Ba}(\mathrm{OH})_{2}}{\text { equivalent mass of } \mathrm{Ba}(\mathrm{OH})_{2}}\)

Molar mass of Ba(OH)₂ = 137 + 2 × 16 + 2 × 1
= 137 + 32 + 2 = 171

Equivalent mass of Ba(OH)₂ = \(\frac{\text { molar mass }}{\text { acidity }}\)
= \(\frac{171}{2}\) = 85.5

∴ ‘n’ factor for Ba(OH)₂ = \(\frac{171}{85.5}\) = 2

(iii) K₂Cr₂O₇
Answer:
Molar mass of K₂Cr₂O₇ = 2 × 39 + 2 × 52 + 7 × 16
= 78 + 104 + 112 = 294

In acid medium Cr₂O₇^-2 is rediced to Cr⁺³
Cr₂O₇^-2 + 6e + 14H⁺ → 2Cr⁺³ + 7H₂O

Equivalent mass of K₂Cr₂O₇ = \(\frac{\text { molar mass }}{\text { No. of electrons taken up }}\)
= \(\frac{294}{6}\) = 49
∴ ‘n’ factor = \(\frac{294}{49}\) = 6 eq. mol^-1.

Question 30.
Calculate the equivalent mass of the following:
(i) Zn
(ii) CO₃^-2
(iii) NO₃^–
(iv) H₃PO₄
Answer:
(i) For elements:
Equivalent mass = \(\frac{\text { Atomic mass }}{\text { valency }}\)
= \(\frac{63.4}{2}\) = 31.7 g eq^-1

(ii) For ions:
Equivalent mass = (sum of the atomic masses of atoms present in the ion) / charge on the ion
For CO₃^-2 ion
Equivalent mass = (1 × atomic mass of ‘N’ + 3 × atomic mass of ‘O’) / 2
= \(\frac{12+3 \times 16}{2}=\frac{60}{2}\) = 30 g eq^-1

(iii) For NO₃^– ion equivalent mass:
= (1 × atomic mass of ‘N’ + 3 × atomic mass of ‘O’) / 1
= \(\frac{14+3 \times 16}{1}\) = 62 g eq^-1

(iv) For acids:
Answer:

Question 31.
0.456g of a metal gives 0.606g of its chloride. Calculate its equivalent mass.
Answer:
By definition, the equivalent mass of a metal is the number of parts by weight of the element that will combine or displace 35.5 parts by weight of chlorine
Weight of metal chloride = 0.606 g
Weight of the metal formed = 0.456 g
∴ Weight of chlorine = 0.150 g
0.150 g of chlorine combines with = 0.606 g of the metal
35.5 g of chlorine combines with = \(\frac{0.606}{0.150}\) × 35.5 g = 143.42 g
∴ Equivalent mass = 143.42 g eq^-1

Question 32.
1.05 g of the metal gave an oxidation 1.5 g of its oxide. Calculate the equivalent mass.
Answer:
By definition, the equivalent mass of a metal is the mass of the metal that will combine or displace 8g of oxygen (equivalent mass of oxygen = 8 g eq^-1)
Weight of metal oxide = 1.5 g
Weight of the metal = 1.05 g
Weight of oxygen = 0.45g
0.45 g of oxygen combines with = 1.05 g of the metal
8.0 g of oxygen will combines with = \(\frac{1.05}{0.45}\) × 8.0 g = 18.7 g
The equivalent mass of the metal is 18.7 g eq^-1.

Question 33.
Calculate the equivalent masses of the following:
(i) HCl
Answer:
Molar mass of HC1 = 1 + 35.5 = 36.5
Basicity of HCl = 1
Equivalent mass of HCl = \(\frac{36.5}{1}\) = 36.5 g eq^-1.

(ii) HNO₃
Answer:
Basicity of HNO₃ = 1 + 14 + 3 × 16 = 63
∴ Equivalent mass of HNO₃ = \(\frac{63}{1}\) = 63 g eq^-1.

(iii) Oxalic acid H₂C₂O₄
Molar mass of oxalic acid (H₂C₂O₄) = 2 × 1 + 2 × 12 + 4 ×16 = 90 Basicity of oxalic acid = 2
∴ Equivalent mass = \(\frac{90}{2}\) = 45 g eq^-1

(iv) Crystalline oxalic acid, H₂C₂O₄ . 2H₂O
Molar mass of crystalline Oxalic add (H₂C₂O₄ . 2H₂O) = Molar mass of oxalic acid + 2 × molar mass of H₂O)
= 90 + 2 × 18 = 126
Basicity = 2
∴ Equivalent mass of crystalline oxalic acid = \(\frac{126}{2}\) = 63 g eq^-1

(v) Ferrous sulphate
Answer:
Molar mass of ferrous sulphate = atomic mass of ‘Fe’ + atomic mass of ‘S’ + 4 × atomic mass of ‘O’
= 56 + 32 + 4 × 16 = 152
In reactions, Fe⁺² ion is oxidised to Fe⁺³ ion
Fe⁺²→ Fe⁺³ + e
∴ ‘n’ factor = 1 (no. of electron involved in half reaction)
∴ Equivalent mass = \(\frac{\text { Molar mass }}{\text { No. of } e^{-\text {involved }}}\)
= \(\frac{152}{1}\)
= 152 g eq^-1.

Question 34.
Calculate the equivalent mass of the following:

(i) Aluminium hydroxide
Answer:
Molar mass of Al(OH)₃ = Atomic mass of ‘Al’ + 3 × Atomic mass of ‘O’ + 3 × Atomic mass of ‘H’
= 27 + 3 + 16 + 3 × 1 = 78
Acidity = 1
Equivalent mass of Al(OH)₃ = \(\frac{\text { Molar mass }}{\text { acidity }}\)
= \(\frac{78}{3}\) = 26 g eq^-1

(ii) Sodium hydroxide
Answer:
Molar mass of NaOH = 23 + 16+ 1 = 40
Acidity = 1
∴ Equivalent mass = \(\frac{\text { Molar mass }}{\text { acidity }}\)
= \(\frac{40}{1}\) = 40 g eq^-1

(iii) Potassium sulphate
Answer:
Molar mass of K₂SO₄ = 2 × Atomic mass of ‘K’ + 1 × Atomic mass of ‘S’ + 4 × Atomic mass of ‘O’
= 2 × 40 + 1 × 32 + 4 × 16
= 80 + 32 + 64 = 176
∴ Equivalent mass odf salt = (\(\left(\frac{\text { atomic mass of cation }}{\text { valency }}\right)\) + \(\left(\begin{array}{c}
\text { sum of atomic equivalent } \\
\text { mass of anion } \\
\hline \text { valency }
\end{array}\right)\))

= \(\frac{40}{1}\) + \(\frac{32+64}{2}\)
= 40 + 48 = 88 g eq^-1.

Question 35.
Define
(i) Empirical formula of a compound,
(ii) Molecular formula of a compound.
Answer:
(i) Empirical formula:
It is the simplest formula of a compound which gives the ratio of the number of different atoms present in one molecule of the compound.

(ii) Molecular formula:
Molecular formula of a compound gives the actual number of different atoms present in one molecule.

Question 36.
3.24g of titanium reacts with oxygen to form 5.40g of the metal oxide. Find the empirical formula of the compound?
Answer:
Step 1:
To find percentage composition of titanium and oxygen.
Weight of the metal oxide = 5.40 g
Weight of the metal = 3.24 g .
Weight of oxygen = 2.16 g
5.40 g of metal oxide contains = 3.24 g of metal
100 g of metal oxide will contains = \(\frac{3.24}{5.40}\) × 100 g = 60 g
∴ Percentage of titanium = 60
Percentage of oxygen = (100 – 60) = 40

Step 2:
To find the empirical formula

Question 37.
A substance an analysis, gave the following percentage composition, Na = 43.4 %, C = 11.3 %, O = 43.3 % calculate its empirical formula.
Answer:

Question 38.
An organic compound contains the following composition by mass: C = 92.3 %; H = 7.7 %. At STP, 10L of the gas has the mass of 11.6g. Find the molecular formula of the compound.

Answer:
The empirical formula is CH.
The empirical formula mass = 12 + 1 = 13
To find the molecular mass:
10 L of the gas at STP weigh = 11.6 g
22.4L of the gas at STP will weigh = \(\frac{1}{12}\) × 22.4 g = 25.98 ≈ 26

n = \(\frac{\text { Molar mass }}{\text { Empirical formula mass }}=\frac{26}{13}\) = 2

Hence, the molecular formula = (Empirical formula)_n
(CH)₂ = C₂H₂
The molecular formula of the compound is C₂H₂.

Question 39.
An organic compound has the following composition by mass: C .= 40.92 %, H = 4.58 %, and the rest oxygen. The molar mass of the compound is 176 g mol^-1. Determine the molecular formula of the compound.
Answer:

The empirical formula is C₃H₄O₃.
Empirical formula mass = 3 × 12 + 4 × 1 + 3 × 16 = 88
Molar mass = 176 g mol^-1
n = \(\frac{\text { Molar mass }}{\text { Empirical formula }}=\frac{176}{88}\) = 2
Molecular formula = (Empirical formula)_n
= (C₃H₄O₃)₂
= C₆H₈O₆

Question 40.
An inorganic compound an analysis gave the following composition. Na = 14.31 %, S = 9.97 %, H = 6.22 %, O = 69.5 %. Assume all the hydrogen atoms are present in combination with oxygen zas water of crystallization. The molar mass of the compound is 322g mol^-1. Find its molecular formula.
Answer:

The Empirical formula is Na₂SH₂₀O₁₄.
Since all hydrogen are present as water of hydration, the number of mole of water present in the compound is 10H₂O.
Hence, the empirical formula Na₂SO₄.10H₂O.
Empirical formula mass of the compound = 2 × 23 + 1 × 32 + 4 × 16 + 10 × 18 = 322.
Molar mass of the compound = 322 g mol^-1
n = \(\frac{\text { Molar mass }}{\text { Empirical formula mass }}\)
= \(\frac{322}{322}\) = 1
Hence the molecular formula is Na₂SO₄ 10H₂O.

Question 41.
What do you understand by the term ‘stoichiometry of the reaction’?
Answer:
Stoichiometry gives the relationship between the amount of reactants consumed and products formed in a balanced chemical equation. It can be
(i) the number of moles of reactants / products
(ii) masses of the reactants / products
(iii) If the reaction is between gases, the volumes of reactants / products represented in a balanced chemical equation.

Question 42.
How many moles of barium sulphate is precipitated when one mole of aluminium sulphate completely reacts with barium chloride?
Answer:
The balanced chemical equation for the reaction is
Al₂(SO₄)₃ + 3BaCl₂ → 3BaSO₄ + 2AlCl₃.
As per stoichiometry, 1 mol of Al₂(SO₄)₃ completes reacts with 3 mol of BaCl₂ and produce 3 mole of barium sulphate. Hence, number of mol of BaSO₄ precipitated is 3.

Question 43.
What is the amount of silver oxide formed when 11.04 g of silver carbonate is strongly heated? Write the balanced chemical equation for the reaction.
Answer:
Ag₂CO₃ → 2AgO + CO₂.
Molar mass og Ag₂CO₃ = 2 × atomic mass of ‘Ag’ + atomic mass of ‘C’ + 3 × atomic mass of ‘O’
= 2 × 108 + 12 + 3 × 16
= 276 u
Molar mass of AgO = atomic mass of ‘Ag’ + atomic mass of ‘O’
= 108 + 16 = 124 g mol^-1

When silver carbonate is heated, carbondioxide leaves as gas. and silver oxide remains as a residue.
According to the equation,
One mole of Ag₂CO₃ = 2 × atomic mass of ‘Ag’ + atomic mass of ‘C’ + 3 × atomic mass of ‘O’
= 2 × 108 + 12 + 3 × 16
= 216 + 12 + 48 = 276 u
Molar mass of AgO = atomic mass of ‘Ag’ + atomic mass of ‘O’
= 108 + 16 = 124 g mol^-1
When silver carbonate is heated, carbondioxide leaves as gas. and silver oxide remains as a residue.
According to the equation,
One mole of Ag₂CO₃ leaves = 2 mol of AgO as residue
276 g of Ag₂O = 124 g of AgO
11.04 of Ag₂O = \(\frac{124}{276}\) × 11.04 g of AgO
= 4.96 g

Question 44.
How much volume of hydrogen is liberated when 0.12g of magnesium reacts with dilute hydrochloric acid?
Answer:
The balanced chemical equation for the reaction is
Mg + 2HCl → MgCl₂ + H₂
Atomic mass of Mg = 24g mol^-1
According to the equation,
1 mol of Mg (24g) will liberate 1 mol of H₂ = 22.4 L at STP
24g of Mg = 22.4 L of H₂ at STP
0.12 g of Mg = \(\frac{22.4}{24}\) × o.12g of H₂ at STP
= 0.112 L or = 112 cm³ at STP

Question 45.
How much volume of nitrogen and hydrogen are required to produce 100cm3 of ammonia?
Answer:
The balanced equation for the reaction is,
N₂ + 3H₂ → 2NH₃
According to the equation,
2 × 22400 cm³ of NH₃ is produced by 1 mol of N₂ = 22400 cm³

Similarly,
2 × 22400 cm³ of NH₃ are produced by = 3 × 22400 cm³ of H₂
100 cm³ of NH₃ will be produced by = \(\frac{3 \times 22400}{2 \times 22400}\) × 100 = 150 cm³

Question 46.
A temperature of about 3000°C is produced when Aluminium reduces ferric oxide and the reaction is given below.
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe
In one process 124 g of aluminium reacted with 601 g of ferric oxide.
(i) Calculate the mass of Al₂O₃ formed.
(ii) How much of the excess reagent is left at the end of the reaction?
Answer:
The balanced equation for the reaction is,
2Al + Fe₂O₃ → Al₂O₃ + 2Fe
(2 x 27) (160) (102)
Number of mol of aluminium taken = \(\frac{\text { mass of Al taken }}{\text { gram atomic weight }}\)
= \(\frac{124}{27}\) = 4.6 mol
No. of mol of Fe₂O₃ taken = \(\frac{\text { mass of } \mathrm{Fe}_{2} \mathrm{O}_{3} \text { taken }}{\text { gram molecular weight }}\)

= \(\frac{601}{160}\) = 3.75 mol
According to equation,
2 mol of al reacts with = 1 mole of Fe₂O₃
4.6 mole of Al will react with = \(\frac{1}{2}\) × 4.6
= 2.3 mol of Fe₂O₃
number of mol of Fe₂O₃ taken = 3.75 mol
number of mol of Fe₂O₃ reacted = 2.30 mol
Hence, Al is the limiting reagent and 1.45 mol of Fe₂O₃ will be left behind, unreacted.

2 mol of Aluminium = 54 g of Al = 102 g of Al₂O₃
124 g of Aluminium = \(\frac{102}{54}\) × 124
= 234 g of Al₂O₃

Mass of Fe₂O₃ left behind = No. of molecular × gram molecular unit
= 1.45 × 160 = 232 g

Question 47.
Explain the term limiting reagent.
Answer:
A limitating reagent is one which is consumed completely in a reaction. The amount of product (i.e., number of moles of product) formed one always determined by the starting moles of limitating reagent.

Question 48.
3 grams of hydrogen reacts with 29g of O₂ to yield H₂O.
(i) Which is the limiting reagent?
2H₂ + O₂ → 2H₂O
2 mol 1 mol 2 mol
Number of mole of H₂ taken = \(\frac{3}{2}\) = 1.5
Number of molel of O₂ taken = \(\frac{29}{32}\) = 0.9.62
∴ 2 mole of H₂ react with 1 mole of O₂
∴ 1.5 mol of H₂ will \(\frac{1}{2}\) × 1.5 = 0.75 mole of O₂
react with \(\frac{1}{2}\) x 1.5
Available mole of O₂ (0.9062) > required mol of O₂ (0.75)
Thus, O₂ is in excess and therefore, H₂ will be the limiting reagent.
(ii) Calculate the maximum amount of H₂O that can be formed.
2 mole of H₂ give 2 mole of H₂O.

(ii) Calculate the amount of one of the reactant which remains unreacted.
2 mole of H₂ react with 1 mole of O₂.
1.5 mole of H₂ will give 1.5 mol of H₂O.

(iii) Calculate the amount of one of the reactant which remains unreacted.
2 mole of H₂ react with 1 mole of O₂.
1.5 mole of H₂ react with 0.75 mole of O₂.

ExcessO₂ = 0.9062 – 0.75 = 0.1562 mol.

Question 49.
Give examples of redox reactions in daily life.
Answer:
(i) When an apple is cut, it turns brown after some time. This is because of oxidation.
(ii) Resting of iron is a redox reaction.
(iii) Oxidation of carbohydrates, lipids etc., to CO₂ and H₂O produce energy in the living organisms.

Question 50.
Calculate the oxidation number of underlined atoms of the following:
(i) KMnO₄
KMnO₄ (Mn in KMnO₄)
Oxidation number of K = 1
Oxidation number of Mn = x
Oxidation number of O = -2
= + 1 + x + 4(-2) = 0
x – 7 = 0
x = + 7 .
Hence the oxidation no.of Mn in KMnO₄ = +7

(ii) K₂Cr₂O₇
Answer:
K₂Cr₂O₇ (Cr in K₂Cr₂O₇)
Oxidation number of K = +1
Oxidation number of Cr = x
Oxidation number of O = -2
2 + 2x – 14 = 0
2x -12 = 0
x = + 6
Hence, the oxidation no. of Cr in K₂Cr₂O₇ + 6

(iii) Na₂S₄O₆
Answer:
Na₂S₄O₆(S in Na₂S₄O₆)
Oxidation number of sodium = 1
Oxidation number of sulphur = x
Oxidation number of Oxygen = -2
i.e. 2 + 4x – 12 = 0
4x – 10 = 0 x = 2.5
This is known as average oxidation state of sulphur. But by definition, oxidation number cannot be fractional. When the compound contains two or more atoms of the same element in different oxidation states chemical bonding method is used to find the oxidation number of elements.

The structure of the compound is

i.e., oxidation number of each sulphur atom linked with the other in the middle is zero, while the oxidation number of each of the remaining two sulphur atom is +5.

(iv) NO₃^–
Answer:
NO₃^– (N in NO₃^–)
Oxidation number of nitrogen = x
Oxidation number of oxygen = -2
x + 3 (- 2) = – 1
x – 6 = – 1
x = + 5
Hence, the oxidation number of nitrogen in NO₃^– is +5.

(v) H₄P₂O₇
Answer:
H₄P₂O₇ (P in H₄P₂O₇ ‘
Oxidation number of hydrogen = + 1
Oxidation number of phosphorous = x
Oxidation number of oxygen = -2
4 + 2x- 14 = 0
2x- 12 = 0
x = + 6
Hence, the Oxidation number of phosphorous in H₄P₂O₇ is + 6.

(vi) ClO₃^–
Answer:
ClO₃^– (Cl in ClO₃^–)
Oxidation number of chlorine = x
Oxidation number of oxygen = -2
x + 3 (- 2) = – 1
x – 6 = – 1 or
x = 5
The oxidation no. of chlorine in ClO₃^– is + 5.

(vii) ASO₃^-2
Answer:
ASO₃^-2 (AS in ASO₃^-2)
Oxidation number of arsenic = x
Oxidation number of oxygen = -2
x + 3 (- 2) = – 2
x – 6 = – 2 (or) x = + 3
Hence, the oxidation number of arsenic in arsenics oxide ASO₃^-2 is + 3.

(viii) S₂O₃^-2
Answer:
S₂O₃^-2 (S in S₂O₃^-2)
Oxidation number of sulphur = x
Oxidation number of oxygen = -2
2x + 3 (- 2) => 2x – 6 = – 2 (or) 2x = 4 or x = +2
Hence, the oxidation number of sulphur is
S₂O₃^-2 is + 2.

(ix) C₆H₁₂O₆
Answer:
(C in C₆H₁₂O₆)
Oxidation number of carbon = x
Oxidation number of hydrogen = +1
Oxidation number of oxygen = -2
6x + 12 – 12 = 0 (or) 6x = 0 (or) x = 0
In organic compounds containing carbon, oxygen and hydrogen, the oxidation number of carbon atom is determined by the formula.
Oxidation number of ‘C’ atom = \(\frac{n_{\mathrm{O}} \times 2-n_{\mathrm{H}}}{n_{\mathrm{C}}}\)

Where n₀ = number of oxygen atoms
n_H = number of hydrogen atoms
n_C = number of carbon atoms

(x) Na₂[Fe(CN)₆]
Na₂[Fe(CN)₆] (Fe in Na₂ [Fe(CN)₆])
This compound contains a cation and a complex anion [Fe(CN)₆]^-4
Let the oxidation number of Iron = x
Each cyanide ion has charge = – 1
The sum of the oxidation number of the metal ion and the charges carried by negative ions should be equal to the charge as the ion. They
x – 6 = -4 or x = + 2.
Hence, Fe is in +2 oxidation state in this compound.

Question 51.
Give examples for the following redox reactions.

(i) Combination
Answer:

(ii) Decomposition
Answer:

(iii) Metal displacement reactions Oxidation
Answer:

(iv) Disproportionation
Answer:

Question 52.
Explain the term competitive electron transfer reaction with an example.
Answer:
When a strip of metallic copper in sliver nitrate solution taken in a beaker and after some time, the solution slowly turns blue. This is due to the formation of Cu²⁺ ions, i.e., copper replaces silver from silver nitrate. The reaction is,

It indicates that between copper and silver, copper has the tendency to release electrons and silver to accept electrons. This type of metal displacement reactions is known as competitive electron transfer reactions.

Question 53.
Balance the following equations by oxidation number method.

(i) CuO + NH₃ → Cu + N₂ + H₂O
Answer:
Step-1: To find atoms undergoing change in oxidation.

Step-2:
To find total increase and decrease in oxidation.
CuO → Cu (decrease of 2 unit per atom)
NH₃ → N₂ (increase of 3 unit per atom)
Total decrease = 2 × 3 = 6
Total increase = 3 × 2 = 6

Step-3:
To balance the total decrease and increase in oxidation, multiply two by 3 and NH₃ by 2. ’ .
3 CuO + 2NH₃ → Cu + N₂ + H₂O

Step-4:
To balance all atoms other than ‘H’ and ‘O’
3 CuO + 2NH₃ → 3Cu + N₂ + H₂O

Step-5:
To balance ‘O’ atoms
3 CuO + 2NH₃ → 3Cu + N₂ + 3H₂O
hydrogen atoms balance by themselves.
Hence the final equation is
3CuO + 2NH₃ → 3Cu + N₂ + 3H₂O

(ii) P + HNO₃ →HPO₃ + NO + H₂O

Step-1:
To find atoms undergoing change in oxidation.

Step-2:
To find total decrease and increase in oxidation.
P → HPO₃ (increase in O.N. of 5 units per atom)
HNO₃ → NO (decrease in O.N. of 3 units per atom)
Total decrease 5 × 3 = 15
Total increase 3 × 5 = 15

Step-3:
To balance the total increase and decrease in the equation, by multiplying P by 3 and HNO₃ by 5.
3P + 5HNO₃ → HPO₃ + NO + H₂O

Step-4:
To balance all atoms other than ‘O’ and ‘H’
3P + 5HNO₃ → 3HPO₃ + 5NO + H₂O

Step-5:
To balance by oxygen atoms Oxygen and hydrogen atoms balance by themselves.
Hence the balanced equation is
3P + 5HNO₃ → 3HPO₃ + 5 NO + H₂O

(iii) KIO₃ + SO₂ + H₂O → KHSO₄ + H₂SO₄ + I₂
Step-1: To find atoms undergoing change in oxidation.

Step-2:
To find the total decrease and increase in oxidation.
KIO₃ → I₂ (decrease of 5 units per atom)
SO₂ → H₂SO₄ (increase of 2 units per atom)
Total decrease = 5 × 2 = 10
Total increase = 2 × 5 = 10

Step-3:
To balance the total decrease and in oxidation increase, multiply KIO₃ by 2 and SO₂ by 5.
2 KIO₃ + 5 SO₂ + H₂O → KHSO₄ + H₂SO₄ + I₂

Step-4:
To balance all atoms other than ‘O’ and ‘H’.
2 KIO₃ + 5 SO₂ + H₂O → 2 KHSO₄ + 3 H₂SO₄ + I₂

Step-5:
To balance ‘O’ atoms 2 KIO₃ + 5 SO₂ + 4H₂O → 2KHSO₄ + 3H₂SO₄ + I₂
Hydrogen atoms balance by themselves. Hence, the balanced equations is
2 KIO₃ + 5SO₂ + 4H₂O → 2KHSO₄ + 3H₂SO₄ + I₂

(iv) C₆H₁₂O₆ + H₂SO₄ → CO₂ + SO₂ + H₂O

Step-1:
To find atoms undergoing change in oxidation.

Step-2:
To find the total increase and decrease in oxidation.
C₆H₁₂O₆ → CO₂ (increase of 4 units per atom
Total increase = 24 units
H₂SO₄ → SO₂ (decrease of 2 units per atom)
Total decrease = 2 x 12 = 24

Step-3:
To balance the total increase and decrease in oxidation, multiply H₂SO₄ by 12.
C₆H₁₂O₆ + 12 H₂SO₄ → CO₂ + SO₂ + H₂O

Step-4:
Balance all atoms other than ‘H’ and ‘O’
C₆H₁₂O₆ + 12 H₂SO₄ → 6CO₂ + 12SO₂ + H₂O

Step-5:
Balance ‘O’ atoms.
C₆H₁₂O₆ + 12 H₂SO₄ → 6CO₂ + 12SO₂ + 18 H₂O
Hydrogen atoms balance by themselves.
Hence the balanced equation is
C₆H₁₂O₆ + 12 H₂SO₄ → 6CO₂ + 12SO₂ + H₂O

(v) C₄H₁₀ + O₂ → CO₂ + H₂O

Step-1:
To find atoms undergoing change in oxidation.

Step-2:
To find the total increase and decrease in oxidation.
C₄H₁₀ → CO₂ + H₂O (decrease by 6.5 units per atom)
– O₂ → H₂O (increases by 2 units per atom)

Step-2:
To find the total increase and decrease
Total decrease = 4 × 6.5 = 26 units in oxidation.
Total increase = 2 × 13 = 26

Step-3:
To balance the total increase and decrease in oxidation multiply C₄H₁₀ by 2 and O₂ by 13.

2C₄H₁₀ + 13O₂ → CO₂ + H₂O

Step-4:
To balance all atoms other ‘H’ and ‘O’.
2C₄H₁₀ + 13O₂ → 8CO₂ + H₂O

Step-5:
To balance ‘O’ and ‘H’ atoms
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
hydrogen atoms balanced by themselves.
Hence balanced equations is
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

(vi) NH₃ + F₂ → HF + N₂
Step – 1:
To find atoms undergoing change in oxidation.

Step-2:
To find the total increase and decrease in oxidation.
NH₃ → N₂ (decrease of 3 units per atom)
F₂ → HF (increase in 1 imit per atom)
Total decrease = 6 units (3 × 2)
Total increase = 6 units (2 × 3)

Step-3:
To balance the increase and decrease in oxidation, multiply NH₃ by 2 and F₂ by 3. 2NH₃ + 3F₂ → HF + N₂

Step-4:
To balance all atoms other than oxygen
2NH₃ + 3F₂ → 6HF + N₂
Hydrogen atom balanced by themselves.
Hence, the balanced equation is
2N₃ + 3F₂ → 6HF + N₂

Question 54.
Balance the following equations by oxidation number method.

(i) K₂Cr₂O₇ + HI → KI + CrI₃ + H₂O + I₂
Answer:
Step-1:
To find atoms undergoing change in oxidation.
K₂Cr₂O₇ + HI → KI + CrI₃ + H₂O + I₂

Step-2:
To find the total increase and decrease in oxidation.
K₂Cr₂O₇ + CrI₃ (decrease of 3 unit / atom = Total decrease = 6 units / 2 atom)
HI → I₂ (increase of 1 unit / atom = total increase 1 × 6 = 6)

Step-3:
To balance the total increase and decrease in oxidation, multiply HI by 6.
K₂Cr₂O₇ + 6 HI → KI + CrI₃ + H₂O + I₂

Step-4:
To balance all atoms other than ‘O’ and ‘H’
K₂Cr₂O₇ + 6 HI → 2KI + 2CrI₃ + H₂O + 3I₂
This makes 14 iodine atoms on RHS. (These iodide ions do not undergo any change in oxidation). Hence to balance the iodine atoms

add 8HI to LHS

i.e.,K₂Cr₂O₇ + 14 HI → 2KI + 2CrI₃ + H₂O + 3I₂
These, the oxygen atoms are balanced by making 7H₂O as RHS.
K₂Cr₂O₇ + 14HI → 2KI + 2CrI₃ + 7H₂O + 3I₂
The hydrogen atoms are balanced by themselves.
Hence the balanced equation is
K₂Cr₂O₇ + 14HI → 2KI + 2CrI₃ + 7H₂O + 3I₂

(ii) Zn + HNO₃ → Zn(NO₃)₂ + N₂O + H₂O.

Step-1:
To find out atoms undergoing change in oxidation.

Step-2:
To find out the total increase and decrease in O.N.

Total increase 2 × 4 = 8
Total decrease 4 × 2 = 8

Step-3:
To balance the total increase and decrease in oxidation.

4Zn + 2HNO₃ → Zn(NO₃)₂ + N₂O + 5H₂O

Step-4:
Balance the equation
4Zn + 2HNO₃ → 4Zn(NO₃)₂ + N₂O + 5H₂O
‘8’ nitrogen atoms, 12 oxygen atoms ‘8’ hydrogen atoms in RHS to be balanced.
For this add 8 HNO₃ to LHS.
4Zn + 10HNO₃ → 4Zn(NO₃)₂ + N₂O + 5H₂O
The equation is now balanced.

(iii) Ag + HNO₃ → AgNO₃ + H₂O + NO
Answer:
Step-1:
To balance atoms undergoing change in oxidation.

Step-2:
To find out total increase and decrease in oxidation.

Total increase =1 × 3 = 3
Total decrease = 3 × 1 = 3

Step-3:
To balance the total increase and decrease in oxidation, multiply Ag by 3.
3Ag + HNO₃ → AgNO₃ + H₂O + NO

Step-4:
To balance all atoms other than hydrogen and oxygen.
3 Ag + 4HNO₃ → 3 AgNO₃ + H₂O + NO

Step-5:
To balance oxygen atoms
3 Ag + 4HNO₃ → 3 AgNO₃ + 2H₂O + NO
Hydrogen atoms balanced by themselves.
The balanced equation is ,
3Ag + 4HNO₃ → 3AgNO₃ + 2H₂O+ NO

(iv) K₂Cr₂O₇ + FeSO₄ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + Fe₂(SO₄)₃ + H₂O
Answer:
Step-1:
To find out atoms undergoing change in oxidation.

Step-2:
To find out total increase and decrease in oxidation.
+6             +3
K₂Cr₂O₇ → Cr₂(SO₄)₃ (Decrease of 3 unit / atom ; 6 unit / 2 atoms)

Total decrease = 6 units
+2             +3
Fe₂(SO₄) → Fe₂(SO₄)₃ (increase of 1 unit / atom)

Total increase 1 × 6 = 6

Step-3:
To balance the total increase and total decrease, multiply FeSO₄ by 6.
K₂Cr₂O₇ + 6FeSO₄ + H₂SO₄ → K₂SO₄ + Fe₂(SO₄)₃ + Cr₂(SO₄)₃ + H₂O

Step-4:
Balance all atoms other than ‘H’ and ‘O’.
Since Fe is oxidised (LHS) to Fe⁺³ (RHS) balance Fe₂(SO₄)₃ by multiplying by 3.
Now the equation becomes,
K₂Cr₂O₇ + 6FeSO₄ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3 Fe₂(SO₄)₃ + H₂O
The sulphur atoms in SO₄^-2 radical does not undergo any change in O.N. There are 7, SO₂^-2 radicals in LHS and 13, SO₄^-2 radicals in RHS. To balance them, multiply H₂SO₄ in LHS by 7.

The equation now becomes
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3 Fe₂(SO₄)₃ + 7H₂O
To balance, ‘H’ atoms multiply H₂O molecules in RHS by 7.
K₂Cr₂O₇ + 6FeSO₄ + 7H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + 3 Fe₂(SO₄)₃ + 7H₂O
This is the balanced equation.

(v) KBr + MnO₂ + H₂SO₄ → KHSO₄ + MnSO₄ + H₂O + Br₂
Answer:
Step-1:
To find atoms undergoing change in oxidation.

Step-2:
To find the total increase and decrease
+4         +2
MnO₄ → MnSO₄ (Decrease in O.N. of 2 units per atom)

KBr^-1 Br₂ (increase in O.N. of 1 unit per atom)
Total decrease = 2 × 1 = 2
Total increase = 1 × 2 = 2

Step-3:
To balance the total increase and decrease in oxidation, multiply KBr by 2.
2KBr + MnO₂ + H₂SO₄ → KHSO₄ + MnSO₄ + H₂O + Br₂

Step-4:
To balance all atoms other than hydrogen and oxygen atoms
2KBr + MnO₂ + H₂SO₄ → 2KHSO₄ + MnSO₄ + H₂O + Br₂

Since SO₄^-2 (sulphate) radical does not undergo any change in oxidation.
balance them RHS = 3 (SO₄^-2) radical .
RHS = 1 (SO₄^-2)

Hence multiply H₂SO₄ in LHS by 3. The equation now becomes
2KBr + MnO₂+ 3H₂SO₄ → 2KHSO₄ + MnSO₄ + H₂O + Br₂

To balance ‘O’ atoms, multiply H₂O in RHS by 2.
2KBr + MnO₂ + 3H₂SO₄ → 2KHSO₄ + MnSO₄ + 2H₂O + Br₂
This is the balanced equation.

(vi) KMnO₄ + KOH + H₂O₂ → K₂MnO₄ + O₂ + H₂O
Answer:
Step-1:
To find out atoms undergoing change in oxidation.

Step-2:
To find out total increase and decrease in oxidation.
+ 7              +6
KMnO₄ → K₂MnO₄ (decrease of 1 unit per atom)
-1            0
H₂O₂ → O₂ (increase of 1 unit per atom or 2 units per atom)
Total decrease = 2

Step-3:
To balance the total increase and decrease in 0.N, multiply KMnO₄ by 2.
2KMnO₄ + KOH + H₂O₂ → K₂MnO₄ + O₂ + H₂O

Step-4:
To balance all atoms other than ‘H’ and ‘O’
2KMnO₄ + 2KOH + H₂O₂ → 2K₂MnO₄ + O₂ + H₂O
number of oxygen atoms on LHS = 12
number of oxygen atoms on RHS = 11
Hence, multiply H₂O in RHS by 2. The equation becomes,
2KMnO₄ + 2KOH + H₂O₂ → 2K₂MnO₄ + O₂ + H₂O
Hydrogen atoms balance by themselves.
The balanced equation is
2KMnO₄ + 2KOH + H₂O₂ → 2KMnO₄ + O₂ + 2H₂O

Question 55.
Balance the following equation by ion electron method.

(i) Cr₂O₇^2- + Fe²⁺ → Cr³⁺ + Fe²⁺ in acidic medium.
Answer:
Cr₂O₇^-2 → Cr⁺³ (reduction) (The O.N of Cr decreases from +6 to +3)
Fe⁺² → Fe⁺³ (oxidation) (The O.N of Fe increases from +2 to +3)
Balancing oxidation and reduction half reactions separately.
Cr₂O₇^-2 → Cr⁺³
Fe⁺² → Fe⁺³

Step-1:
Balance all atoms, other than hydrogen and oxygen.
Cr₂O₇^-2 → 2Cr⁺³
Fe⁺² → Fe⁺³

Step-2:
Balance oxygen atom by adding equal number of H₂O molecules on the side deficient of oxygen.
Cr₂O₇^-2 → 2Cr⁺³ + H₂O
Fe⁺² → Fe⁺³.

Step-3:
To balance, hydrogen atom in acidic medium, add equal number of H⁺ ions on the side deficient of hydrogen atom.
14H⁺ + Cr₂O₇^-2 → 2Cr⁺³ + 7H₂O
Fe⁺² → Fe⁺³

Step-4:
Balance the charges in such a way the charges on either side be the same,
6e + Cr₂O₇^-2+ 14H⁺ → 2Cr⁺³ + 7H₂O
Fe⁺² → Fe⁺³
Add both equations in such a way the number of electrons taken up by the oxidising agent and the number of electron given by the reducing agent be the same,

(ii) Cr₂O₇^-2 + C₂H₄O → Cr³⁺ + C₂H₄O₂ in acidic medium.
Answer:
Cr₂O₇^-2 → Cr⁺³ (Reduction)
C₂H₄O → C₂H₄O₂ (Oxidation)

Step-1:
Balance all atoms other than oxygen
Cr₂O₇^-2 → 2Cr⁺³
C₂H₄O → C₂H₄O₂

Step-2:
Balance oxygen atom by adding water molecules on the side deficient of oxygen atom.
Cr₂O₇^-2 → 2Cr⁺³ + H₂O
C₂H₄O + H₂O → C₂H₄O₂

Step-3:
To balance, hydrogen atom in acidic medium, add equal number of H⁺ ions on the side deficient of hydrogen atom.

Cr₂O₇^-2 + 14 H^-1 → 2Cr⁺³ + H₂O …………….(1)
C₂H₄O + H₂O → C₂H₄O₂ + 2H⁺ + 2e ……….(2)

Multiply equation (2) by 3 and add with equation (1)

Cr₂O₇^-2 + 14 H⁺ + 6e → 2 Cr⁺³ + H₂O
C₂H₄O + H₂O → C₂H₄O₂ + 2H⁺ + 2e × 3
Cr₂O₇^-2 + C₂H₄O + 8H⁺ → 2 Cr⁺³ + 4 H₂O + 3 C₂H₄O₂

(iii) Zn + NO₃^– → Zn⁺² + NH₄ in basic medium.
Zn → Zn⁺² + 2e (oxidation)
NO₃^– → NH₄⁺ (Reduction)
Answer:
Step-1:
Balance all atoms other than hydrogen and oxygen.
Zn → Zn⁺²
NO₃^– → NH₄⁺

Step-2:
For balancing oxygen atom in alkaline medium, add H₂O molecules on the side deficient of oxygen atom. Then balance hydrogen atom, add H₂O molecule to file side deficient in hydrogen and equal number of ,OH^– on the other side.

Zn → Zn⁺²
NO₃^– → NH₄⁺ + 3H₂O
NO₃^– + 10H₂O → NH₄⁺ + 3H₃O
NO₃^– + 10H₃O → NH₄⁺ + 3H₂O + 10 OH^–

Step-3:
Add electrons to balance charge
Zn → Zn⁺² + 2e
NO₃^– + 7H₂O + 8e → NH₄⁺ + 10 OH^–

This is the balanced equation.

Question 56.
The approximate production of Na₂CO₃ per month is 424 × 10⁶ g while that of methyl alcohol is 320 × 10⁶ g. Which is produced more in terms of moles?
Answer:

Gram molecular weight of methyl alcohol (CH₃OH) = 32g
number of mol of methyl alcohol = \(\frac{320 \times 10^{6}}{32}\)
= 10 × 10⁶ mol.
Hence methyl alcohol is produced move in terms of moles.

Question 57.
Calculate the number of moles of carbon atoms in three moles of ethane.
Answer:
Ethane = Molecular formula C₂H₆
1 mol of ethane = 2 mol of carbon atoms
3 mol of ethane = 2 × 3 mol of carbon atoms = 6 mol.

Question 58.
Find the molecular mass of FeSO₄7H₂O.
Answer:
Molecular mass is the sum of atomic masses of various atoms present in one molecule of the compound.
FeSO₄7H₂O:

Atomic mass of Fe = 55.85
Atomic mass of S = 32
Atomic mass of O = 16
Atomic mass of H = 1
Molecular mass of 7H₂O = 7 × 18 = 126
Molecular mass of FeSO₄.7H₂O = 55.85 + 32 + 4 × 16 + 126
= 277.854 = 278 u

Question 59.
Mass of one atom of an element is 6.66 × 10^-23 g. How many moles of element are there in 0.320kg?
Answer:
No. of mol = (Weight of the substance in gram) / Gram molecular mass.

Step -1:
Find the gram molecular weight of the atom.

Mass of 6.023 × 10²³ atoms = 1 mole = 1 gram molecular weight
Mass of one atom of the element = 6.626 × 10^-23 g
Mass of 6.023 × 10²³ atoms = 6.626 × 10^-23 × 6.023 × 10²³ = 40.1 .
Therefore, the gram molecular weight of the element is 40.1 g

Step -2:
To find the number of. mol in 0.320 kg
Number of mol = \(\frac{0.320 \times 1000 g}{40.1 g}\)
= 7.98 ≈ 8 mol

Question 60.
The density of CO₂ = 1.977kgm^-3 at STP. Calculate the molecular mass of CO₂.
Answer:
Density of CO₂ = 1.977 kg m^-3
This means 1m³ of CO₂ weighs 1.977 kg.
22.4 L (or) 22.4 × 10^-3 m^-3 = 1.977 × 22.4 × 10^-3 kg
Weigh = 1.977 × 22.4 × 10^-3 × 10³ g
= 44.28g = 44.28 u.
The molecular mass of CO₂ is the mass of 22.4 L or 22.4 × 10~3 m³ of CO₂.

Question 61.
How many moles of glucose are present in 720 g of glucose?
Answer:
The Molecular formula of glucose: C₆H₁₂O₆
Molecular mass of glucose = 6 × 12 + 12 + 6 × 16] = 72 + 12 + 96 = 180M.
180 g of glucose = 1 mol
720 g of glucose = \(\frac{1 \times 720}{180}\) = 4 mol

Question 62.
Calculate the weight of 0.2 mole of sodium carbonate.
Answer:
Moleular formula of sodium carbonate = Na₂CO₃
Molecular mass of sodium carbonate = 2 × 23 + 12 + 3× 16
= 46 + 12 + 48 = 106 u
1 mol of Na₂CO₃ = 106 grams
0.2 mol of Na₂CO₃ = 106 × 0.2 gram = 21.2g

Question 63.
Calculate the equivalent mass of bicarbonate ion.
Answer:
Equivalent mass of an ion = (sum of atomic mass of constituent atom in an ion) / charge on the ion
Bi-carbonate ion has the formula HCO₃^–.
Equivalent mass ofl carbonate ion = (atomic mass of hydrogen atom + atomic mass of carbon atom + 3 × atomic mass of oxygen atom) / (charge on bi – carbonate ion)
= \(\frac{1+12+3 \times 16}{1}=\frac{1+12+48}{1}\) = 61
The equivalent mass of carbonate ion is 61 g. eq^-1

Question 64.
Calculate the equivalent mass of barium hydroxide.
Answer:
Barium hydroxide has the formula Ba(OH)₂.
Molecular mass of barium hydroxide = atomic mass of Ba atom + 2 × atomic mass of oxygen atom + 2 × atomic mass of hydrogen atom
= 137.27 + 2 × 16 + 2 × 1
= 137.27 + 32 + 2 = 171.27
Equivalent mass of barium hydroxide = (Molecular mass) / (No. of ‘OH’ group present)
= \(\frac{171.27}{2}\) = 85.64 g. eq^-1.

Question 65.
Calculate the equivalent mass of hydrated sodium carbonate.
Answer:
Hydrated sodium carbonate has the formula Na₂CO₃.10H₂O
Molecular mass of Na₂CO₃.10H₂O = 2 × atomic mass of Na + 1 × atomic mass of C + 3 × atomic mass of oxygen + 10 × molar mass of H₂O
= 2 × 23 + 12 + 3 × 16 + 10 × 18 = 46+ 12 + 48 + 180 = 286u
2 NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O
Equivalent mass of NaOH = Molar mass of NaOH
2 equivalent of NaOH = 1 equivalent of Na₂CO₃
1 equivalent of NaOH = \(\frac{1}{2}\) equivalent mass of Na₂CO₃

Equivalent mass = \(\frac{\text { Molar mass }}{2}\)
= \(\frac{286}{2}\) = 143 g eq^-1

Question 66.
Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?
Answer:
Molecular formula of boric acid = H₃BO₃
Molecular mass of boric acid = 3 × atomic mass of hydrogen atom + atomic mass of boron atom + 3 × atomic mass of oxygen atom
= 3 × 1 + 10.8 + 3 × 10
= 3 + 10.8 + 48 = 80.4u.
1 mol of boric acid = 80.4 gram
0.543 mol = \(\frac{80.4}{1}\) = 43.6g
Remember the formula:
Number of mol of a substance = (Weight of the sample in gram) / gram molecular mass

Question 67.
A compound contains 50 % of X (atomic mass 10) and 50 % Y (atomic mass 20).Give its molecular formula.
Answer:

Empirical formula = XY.
Molecular formula can be calculated one of the molecular mass of the compound is given by using the formula.
Molecular mass = (Empirical formula mass)_n
Molecular formula = (Empirical formula)_n
Where n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}\)
If the molecular mass is assured to be equal to 20. Then, the molecular formula of the compound is XY.

Question 68.
Calculate the mass of sodium (in kg) present in 95 kg of a crude sample of sodium nitrate whose percentage purity is 70%.
Answer:
Mass of impure sample of sodim nitrate (NaNO₃) = 95 × 1000 g
Mass of pure sample of sodium nitrate (NaNO₃) = 95 × 1000 × \(\frac{70}{100}\)
= 66500 g
1 mole of pure NaNO₃ = 85 g
85 g NaNO₃ = 23 g of Na
66,500 of NaNO₃ = \(\frac{23}{85}\) × 66500 g of NaNO₃
= 17994.1 gram = 17.99 kg

Choose the correct answer:

Question 1.
Which one of the following is the standard for atomic mass?
(a) ₁H¹
(b) ₆C¹²
(c) ₆C¹⁴
(d) ₈O¹⁶
Answer:
(b)
Hint:
If ₁H¹ is used as a reference, the atomic mass is expressed in hydrogen scale and ₈O¹⁶ is used as a reference the atomic mass is expressed in oxygen scale ₆C¹² is used as a standard atom and atomic mass is expressed in C-12- scale.

Question 2.
One mole of CO₂ contains:
(a) 6.023 × 10²³ atoms of C
(b) 6.023 × 10²³ atoms of O
(c) 18.1 × 10²³ molecules of CO₂
(d) 3g atoms of CO₂
Answer: (a)
Hint: 1 mole of CO₂ = 1 mole of C = 2 mole of ‘O’ atoms
= 6.023 × 10²³ atoms of carbon
= 2 × 6.023 × 10²³ atoms of oxygen
= 6.023 × 10²³ molecules of CO₂

Question 3.
The largest number of molecules is in:
(a) 54 g of nitrogen pentoxide
(b) 28 g of carbon dioxide
(c) 36 g of water
(d) 46 g of ethyl alcohol
Answer: (a)
Hint:
N₂O₅: Molar mass =108
108 g N₂O₅ = 1 mol
= 6.023 × 10²³ molecules.
54g of N₂O₅ = \(\frac{6.023 \times 10^{23}}{108}\) × 54
= 3.012 × 10²³ molecules

CO₂: Molar mass = 44.
44g of CO₂ = 1 mol
= 6.023 × 10²³ molecules
28g of CO₂ = \(\frac{6.023 \times 10^{23}}{44}\) × 28
= 0.636 × 6.023 × 10²³ mol

H₂O: Molar mass = 18.
18g of H₂O – 1 mol = 6.023 × 10²³ molecules
36g H₂O = 2mol
= 2 × 6.023 × 10²³ molecules

C2H5 OH: Molar mass = 46
46g = 1 mol
= 6.023 × 10²³ molecules

Question 4.
The number of moles of H₂ in 0.224 litre of hydrogen gas at STP is:
(a) 1
(b) 0.1
(c) 0.01
(d) 0.001
Answer: (c)
Hint: 22.4 L of H₂ at STP = 1 mol of H₂O
0.224 L of H₂ at STP = \(\frac{1 \times 0.224}{22.4}\) = 0.01 mol

Question 5.
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be:
(a) 3 mol
(b) 4 mol
(c) 1 mol
(d) 2 mol
Answer: (b)
Hint:
2H₂ + O₂ → 2H₂O
10 g of H₂ = \(\frac{10}{2}\) = 5 mol of H₂
64 g of O₂= \(\frac{1}{12}\) = 2 mol of O₂
1 mol of O₂ = 2 mol of H₂ as per equation .
2 mol of O₂ = 4 mol of H₂
i.e., one mol of H₂ remain excess.
O₂ of the limiting reagent
1 mol O₂ = 2 mol H₂O
2 mol O₂ = 4 mol H₂O

Question 6.
6.023 × 10²⁰ molecules of urea are present in 100 ml of its Solution. The concentration of the solution is:
(a) 0.02 M
(b) 0.1 M
(c) 0.01 M
(d) 0.001 M
Answer: (c)
Hint: Molarity is defined by the number of mol of solute present in 1L of the solutions.
Molarity = \(\frac{\text { no. of mole of the solute } \times 1000}{\text { Volume of solution in } \mathrm{cm}^{3}}\)
6.023 × 10²³ molecule of urea = 1 mol
6.023 × 10²⁰ molecules of urea
= \(\frac{1 \times 6.023 \times 10^{20}}{6.023 \times 10^{23}}\)
= 10^-3 mol
∴ molarity = \(\frac{10^{-3} \times 1000}{100}\) = 0.01 M

Question 7.
Two containers A and B of equal volume contain 6g of each O₂ and SO₂ at 300K and latm. Then:
(a) No. of molecules in A is less than that in B
(b) No.of molecules in A is more than that in B
(c) No. of molecules in A and B are same
(d) none of these
Answer: (b)
Hint: 6g O₂ = \(\frac{6}{32}\) =0.1875 mol
6g SO₂ = \(\frac{6}{64}\) = 0.0934 mol
At a given temperature and pressure, number of mole is proportional to number of molecule.
Here (b) is the correct option.

Question 8.
The number of molecules in 16g of methane is:
(a) 3.023 × 10²³
(b) 6.023 × 10²³
(c) 16/6.023 × 10²³
(d) 6.023/3 × 10²³
Answer: (b)
Hint: CH₄:
Molar mass = 16g mol^-1
16g methane = 1 mol
= 6.023 × 10²³ molecules

Question 9.
No. of atoms in 4.25g of ammonia is:
(a) 1 × 10²³
(b) 2 × 10²³
(c) 4 × 10²³
(d) 6 × 10²³
Answer: (d)
Hint:
1 mol of NH₃ = (3 mol of hydrogen + 1 mol of nitrogen) = 4 mol of atoms
4.25g of NH₃ = \(\frac{4.25}{17}\) = 0.25 mol of N₃
1 mol of NH₃ = 4 × 6.023 × 10²³ atom
0.25 mol NH₃ = \(\frac{4 \times 6.023 \times 10^{23}}{1}\) × 0.25
= 6.023 × 10²³ atom

Question 10.
The number of molecules in a drop of water (0.0018mL) at room temperature is: •
(a) 6.02 × 10¹⁹
(b) 1.084 × 10¹⁸
(c) 4.84 × 10¹⁷
(d) 6.02 × 10²³
Answer: (a)
Hint:
0.0018ml of H₂O = 0.0018g H₂O
(density of H₂O) = 1g / cc
18g H₂O = 1 mol
= 6.023 × 10²³ molecules
0.0018g of H₂O = \(\frac{6.023 \times 10^{-23} \times 0.0018}{18}\)
= 6.023 × 10²³ × 10^-4
= 6.023 ×10¹⁹ molecule

Question 11.
7.5g of a gas occupy 5.6 litres of volume at STP. The gas:
(a) NO
(b) N₂O
(c) CO
(d) CO₂
Answer: (a)
Hint: 1 mol of gas at NTP will occupy 22.4L at 1 mol is molecular weight expressed in grams
5.6L of gas at STP = 75 g
22.4L of the gas at STP = \(\frac{7.5}{5.6}\) × 22.4 g = 30 g
i.e. The molecular weight of the gas is 30.
NO = Molar mass = 30
N₂O = Molar mass = 44
CO = Molar mass = 28
CO₂ = Molar mass = 44

Question 12.
The mass in grams of 0.45 mole of Ca²⁺ ions:
(a) 1.8
(b) 40
(c) 36
(d) 18
Answer: (d)
Hint: Ca and Ca⁺ ion have the same atomic mass, i.e. 40.
1 mol of Ca⁺² = 40g
0.45 mol of Ca⁺² = 40 × 0.45 = 18g

Question 13.
The mass of one molecule of HI in grams is:
(a) 2.125 × 10^-22
(b) 128
(c) 127
(d) 6.02 × 10^-23
Answer: (a)
Hint: Molecular mass of HI = 1 + 127 = 128
6.023 × 10²³ molecules of HI = 128 g
1 molecule of HI = \(\frac{128}{6.023 \times 10_{4}^{23}}\)
= 21.25 × 10^-23
= 2.125 × 10^-23

Question 14.
Avogadro’s number is the number of molecules present in:
(a) 1 g of molecule
(b) 1 g atom of molecule
(c) gram molecular mass
(d) 1 lit of molecule
Answer: (c)
Hint: Gram molecular mass is the molecular mass expressed in grams which correspond to one mol of the molecule. The number of molecules present in one mole of the substance is known as the Avogadro number.

Question 15.
Which of the following contains same number of carbon atoms as are in 6.0 g of carbon (C – 12)?
(a) 6.0 g ethane
(b) 8.0 g methane
(c) 21.0 g Propane
(d) 28.0 g CO
Answer: (b)
Hint: 6g of carbon = \(\frac{6}{12}\) = 0.5 mol.
The given choice should contain 0.5 mol of carbon atoms.
6g of ethane (C₂H₆) = \(\frac{6}{30}\) = 0.2 mol

(a) 1 mol of ethane = 2 mol of carbon atoms
0.2 mol = 2 × 0.2 × 6.023 × 10²³ carbon atoms
= 0.4 × 6.023 × 10²³ carbon atoms

(b) 1 mole of methane = 1 mol of carbon atom
8g of methane = \(\frac{8}{16}\) = 0.5 mol
∴ 0.5 mol of methane = 0.5 mol of carbon atom

(c) 21g of propane = \(\frac{21}{44}\) = 0.477 mol.
1 mol of C₃H₈ = 3 mol of carbon atoms
0.477 mol of C₃H₈ = 3 × 0.477
= 1.432 mol of carbon atom

(d) 28g of CO = 1 mol = 1 mol of CO.
Hence 6g of carbon and 8g of methane contain same number of carbon atoms.

Question 16.
Equivalent mass of KMnO₄ when it is converted to MnSO₄ is equal to molar mass divided by:
(a) 6
(b) 4
(c) 5
(d) 2
Answer: (c)
Hint: MnO₄^– → Mn⁺²
MnO₄^– + 8H⁺ + 5e Mn⁺² + 4H₂O
Equivalent mass = (Molar mass) / (no. of e^– involved in half reaction)
= M / 5

Question 17.
How many equivalents of sodium sulphate is formed when sulphuric acid is completely neutralized by a base NaOH:
(a) 0.2
(b) 2
(c) 0.1
(d) 1
Answer: (b)
Hint: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
1 mol of H₂SO₄ = 2 equivalent of H₂SO₄
.’. 2 equivalent of H₂SO₄ will form 2 equivalent of Na₂SO₄
Note:
One equivalent of any substance will always react with one equivalent of any other substance and one equivalent of the product.

Question 18.
Cl₂ changes to Cl^– and ClO^– in cold NaOH. Equivalent mass of Cl₂ will be:
(a) Molar mass/2
(b) Molar mass/1
(c) Molar mass/3
(d) 2 × Molar mass/2
Answer: (a)
Hint: (Reduction)
Cl₂ → Cl^–
Cl₂ → 2Cl^–
Cl₂ + 2e → 2Cl^–

(Oxidation)
Cl₂ → 2OCl^–
Cl₂ + 4 0H^– → 2H₂O + 2OCl^– + 2e
Cl₂ + 2e → 2Cl^–
Cl₂ + 2OH^– → Cl + OCl^– + H₂O

\(\frac{\mathrm{Cl}_{2}+4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{OCl}^{-}+2 e}{2 \mathrm{Cl}_{2}+4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{Cl}^{-}+2 \mathrm{OCl}^{-}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{or})}\)

Since this is a 2 electron transfer reaction equivalent mass of chlorine is \(\frac{M}{2}\).

Question 19.
Equivalent mass of KMnO₄ in acidic medium, concentrated alkaline medium and dilute basic medium respectively are \(\frac{M}{5}\), M, M. Reduced products can be:
(a) MnO₂, MnO₄^–, Mn²⁺
(b) MnO₂, Mn²⁺, MnO₄^2-
(c) Mn²⁺, MnO₂, MnO₄^2-
(d) Mn²⁺, MnO₄^2-, MnO₂
Answer: (d)
Hint: KMnO₄ (MnO₄^–) acts as an oxidising agent in acidic alkaline and in neutral medium. The half reactions are:
MnO₄^– + 8H⁺ 5e → Mn⁺² + 4 H₂O(acidic)
Equivalent mass = \(\frac{\text { Molar mass }}{5}\)

MnO₄^– + e → MnO₄^-2 (alkaline)
Equivalent mass = Molar mass
MnO₄^– + 2H₂O + 3e → MnO₂ + 4 OH^– (neutral or mild alkaline)
Equivalent mass = \(\frac{\text { Molar mass }}{3}\)

Question 20.
The empirical formula of hydrogen peroxide is:
(a) HO
(b) H₂O
(c) H₃O⁺
(d) H₂O₂
Answer: (a)
Hint: The formula of hydrogen peroxide is H₂O₂. Its simplest formula is HO.

Question 21.
Molecular mass =
(a) Vapour Density × 2
(b) Vapour Density ÷ 2
(c) Vapour Density × 3
(d) Vapour Density
Answer: (a)

Question 22.
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage of purity of magnesium carbonate in the sample?
(a) 60
(b) 84
(c) 75
(d) 96
Answer: (b)
Hint: MgCO₃ → MgO + CO₂
24.3 + 12 + 48 = 84.3 g 24.3 + 16 = 40.3 g
40.3 g MgO = 84.3 g MgCO₃
8 g MgO = \(\frac{84.3}{40.3}\) × 8 = 16.7 g
Percentage of purity = \(\frac{16.7}{20}\) × 100 = 83.67 % ≈ 84 %

Question 23.
What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO₃ is mixed with 50 mL of 5.8% NaCl solution?
(a) 7 g
(b) 14 g
(c) 28 g
(d) 35 g
Answer: (a)
Hint:AgNO₃ + NaCl → AgCl + NaNO₃
Molarity of AgNO₃ = \(\frac{\text { weight of } \mathrm{AgNO}_{3}}{\text { gram molecular weight }} \times \frac{1000}{\text { volume in } \mathrm{cm}^{3}} \text {. }\)
= \(\frac{16.9}{170} \times \frac{1000}{50}\) ≈ 2 M

Molarity of NaCl = \(\frac{16.9}{170} \times \frac{1000}{50}\) ≈ 1 M
No. of mol. of AgNO₃ = \(\frac{\mathrm{M} \times \text { Volume in } \mathrm{cm}^{3}}{1000}\)
= \(\frac{2 \times 50}{1000}\) 1 M

No. of mol of NaCl = \(\frac{1 \times 50}{1000}\) = 0.05 M
NaCl is the limiting reagent.
∴ No. of mol of AgCl formed = 0.05 M
Mass of AgCl formed = 0.05 × 143.58 = 7.188 g ≈ 7 g

Question 24.
When 22L of hydrogefo gas is mixed with 11.2 L of chlorine gas, each at STP, the moles of HCl gas formed is equal to:
(a) 2
(b) 0.5
(c) 1.5
(d) 1
Answer: (a)
Hint:. H₂ + Cl₂ → 2HCl
Chlorine is the limiting reagent.
11.2L of chlorine will form 2 mol of HCl as per equation.

Question 25.
5.6 L of a gas at STP are found to have mass of 11 g. The molecular mass of the gas is:
(a) 36
(b) 48
(c) 40
(d) 44
Answer: (d)
Hint: 5.6L of the gas = 11 g
22.4 L of the gas = \(\frac{1}{12}\) × 22.4 = 44g .
Hence the Molar mass of the gas is 44.

Question 26.
Oxidation number of Fluorine in all compounds is:
(a) +1
(b) – 1
(c) 0
(d) – 2
Answer: (b)

Question 27.
In redox reaction which of the following is true?
(a) Number of electrons lost is more than number of electrons gained
(b) Number of electrons lost is less than number of electrons gained
(c) Number of electrons lost is equal number of electrons gained
(d) No transfer and gain of electrons during the reaction.
Answer: (c)

Question 28.
Match the entities in column I with appropriate entities in column II.

Column I	Column II
(No. of moles)	(Amount)
(i) 0.1 mole	(A) 1.5 x 1023 molecules of oxygen gas
(ii) 0.2 mole	(B) 0.1 g atom of iron
(iii) 0.25 mole	(C) 9 ml of water
(iv) 0.5 mole	(D) 4480 ml of CO, at STP

(a) (i) – (B), (ii) – (D), (iii) – (A), (iv) – (C)
(b) (i) – (D), (ii) – (A), (iii) – (C), (iv) – (B)
(c) (i) – (D), (ii) – (A), (iii) – (B), (iv) – (C)
(d) (i) – (C), (ii) – (A), (iii) – (D), (iv) – (B)
Answer: (a)

Question 29.
Match the entities in column I with appropriate entities in column II.

Column I (Mass of the substance)	Column II (Number of molecules)
(i) 54g of N205	(A)3.8 × 10
(ii) 28g of C02	(B) 6.023 × 1023
(iii) 36g of H20	(C)3.012 × l023
(iv) 46g of ethyl alcohol (C2H5OH)	(D) 1.2 × l024

(a) (i) – (B), (ii) – (D), (iii) – (A), (iv) – (C)
(b) (i) – (D), (ii) – (A), (iii) – (C), (iv) – (B)
(c) (i) – (D), (ii) – (A), (iii) – (B), (iv) – (C)
(d) (i) – (C), (ii) – (A), (iii) – (D), (iv) – (B)
Answer: (d)

Question 30.
In the following questions one of the option is different from the rest.
(a) 40 g of NaOH in 1L of H₂O
(b) 98 g of H2S04 in 1L of H₂O
(c) 58.5 g of NaCl in 1L of H₂O
(d) 98% H₂SO₄ by mass, which has a density of 1.84 g cm³.
Answer: (d)
Hint:
All (a),(b),(c) represent a 1 molar solution of that substance while (d) has the molality of 18.4.
98 % H₂SO₄ by mass means 98 g of H₂SO₄ are present in 100 g of the solution.
= \(\frac{100 g}{1.84 g \mathrm{~cm}^{3}}\) = 0.5435 L
Mass of H₂SO₄ = 98 g mol^-1 = 1 mole
Molarity of the given solution = \(\frac{1 \mathrm{~mol}}{0.5345 \mathrm{~L}}\) = 18.4 M

Question 31.
In the following questions one of the option is different from the rest.
(a) Air
(b) Brass
(e) An aqueous solution of sugar
(d) Petrol
Answer: (d)
Hint: While (a),(b),(c) are homogeneous mixture, (d) is a heterogeneous mixture.

Question 32.
From the following, choose the correct pair of diatomic molecules.
(a) Copper and hydrogen
(b) Copper and phosphorus
(c) Nitrogen and oxygen
(d) Oxygen and Sulphur
Answer: (c)
Hint: Copper is mono atomic (Cu), hydrogen is diatomic (H₂), phosphorus and sulphur are poly atomic. N₂ and O₂ are diatomic.

Question 33.
Which of the following pairs of the solutions contain same number of mol of the solute?
(a) 500 ml of 1M solution of NaOH and 1000 ml of 0.5 M solution of NaOH.
(b) 500 ml of 1M solution of NaOH and 100 ml of 0.1 M solution of NaOH.
(c) 500 ml of 0.1 M solution of NaOH and 100 ml of 0.5 M solution of NaOH.
(d) 1000 ml of 0.5M solution of NaOH and 100 ml of 0 .1 M solution of NaOH.
Answer: (c)
Hint:
No. of moles = \(\frac{\mathrm{M} \times \mathrm{vol} \mathrm{in} \mathrm{cm}^{3}}{1000}\)

= \(\frac{0.1 \times 500}{1000}=\frac{50}{1000}\) = 0.05

= \(\frac{0.5 \times 100}{1000}\) = 0.05

Question 34.
Statement 1:
A single C¹² atom has a mass exactly 12 amu and a mole of these atoms has a mass of exactly 12 grams. .
Statement 2:
A mole of atoms of any element has a mass in grams equal to its atomic mass.
(a) Statement (1) is true, Statement (2) is true, Statement (2) is the correct explanation for Statement (1).
(b) Statement (1) is true, Statement (2) is true, Statement (2) is not a correct explanation for Statement (1).
(c) Statement (1) is true, Statement (2) is false.
(d) Statement (1) is false, Statement (2) is true.
Answer: (a)

Question 35.
Statement 1:
In a gaseous reaction, the ratio of volumes of reactants – and gaseous products is in agreement with their molar ratio.
Statement 2:
Volume of a gas is inversely proportional to its moles at a particular temperature and pressure.
(a) Statement (1) is true, Statement (2) is true, Statement (2) is the correct explanation for Statement (1).
(b) Statement (1) is true, Statement (2) is true, Statement (2) is not a correct explanation for Statement (1).
(c) Statement (1) is true, Statement (2) is false.
(d) Statement (1) is false, Statement (2) is true.
Answer: (c)
Hint: Correct Statement (2):
Volume of a gas is directly proportional to its moles at a particular temperature and pressure.

Question 36.
Statement 1:
Both 100 g of sodium carbonate and 12g of carbon have the same number of carbon atoms.
Statement 2:
Both contain lg atom of carbon which contains 6.023 × 10²³ carbon atoms
(a) Statement (1) is true^Statement (2) is true, Statement (2) is the correct explanation for Statement (1).
(b) Statement (1) is true, Statement (2) is true, Statement (2) is not a correct explanation for Statement (1).
(c) Statement (1) is true, Statement (2) is false.
(d) Statement (1) is false, Statement (2) is true.
Answer: (a)
Hint:
106g Na₂CO₃ = 1 mol = 1 g atom of C
12g C = 1 g atom of C

Question 37.
Statement 1:
In a combustion reaction in the air, oxygen is the limiting reagent.
Statement 2:
Oxygen is present in limited amount(only 21 %) in air.
(a) Statement (1) is true, Statement (2) is true, Statement (2) is the correct explanation for Statement (1).
(b) Statement (1) is true, Statement (2) is true, Statement (2) is not a correct explanation for Statement (1).
(c) Statement (1) is true, Statement (2) is false.
(d Statement (1) is false, Statement (2) is true.
Answer: (d)
Hint: Correct Statement (1): In a combustion reaction in air, the substance being burnt is the limiting reagent.
Correct Statement (2): Oxygen is present in large excess in the air and is the excess reagent.

Question 38.
Choose the correct Statements from among the following:
In a reaction, A+B₂ → 2AB
(a) ‘B’ is the limiting reagent when 300 atoms of A and 200 molecules of B react.
(b) ‘B’ is the limiting reagent when 2 mpl of A and 3 mol of B react.
(c) ‘A’ is the limiting reagent when 100 atoms A and 100 molecules of B react.
(d) ‘A’ is the limiting reagent when 5 mol of ’ A reacts with 2.5 mol of B.
Answer: (a)
Hint: According to given reaction 1 atom of A reacts with one molecule of B. Therefore 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be left unreacted. Hence ‘B’ is the limiting reagent and A is the reagent in excess.

Question 39.
Choose the correct Statements from among the following:
(a) All atomic masses of atoms are expressed taking C¹² isotope as the reference and this scale is referred as unified scale (u).
(b) One atomic mass unit (amu) is equal to the mass of one atom of C – 12 isotope.
(c) The atomic mass of an element is the average relative mass of its atoms (isotopes) as compared to \(\frac{1}{12}\)th mass of the atom of C – 12 isotope.
(d) The empirical formula gives the exact number of atoms of all the elements present in that compound.
Answer: (a)
Hint: Correct Statement:
(b) One atomic mass unit (amu) is equal to \(\frac{1}{12}\)th of the mass of an atom of carbon-12 isotope.
(c) The atomic mass is the average relative masses of atoms compared with an atom of carbon-12 taken as 12.
(d) The empirical formula gives the ratio of the number of different atoms present in a molecule of a compound.

Question 40.
Which of the following Statements about the compound is incorrect?
(a) A molecule of a compound has atoms of different elements.
(b) A compound cannot be separated into its constituents by physical methods of separation.
(c) The compound retains the physical properties of its constituents.
(d) The ratio of atoms of different elements in a compound is fixed.
Answer: (c)

Question 41.
Sodium sulphate has the formula Na₂SO₄. Atomic masses of Na, S and oxygen are 23, 32 and 16 respectively. Using these data choose the incorrect Statement among the following.
(a) One mole of sodium sulphate contains two moles of sodium, one mole of sulphur and few moles of oxygen atoms.
(b) One mole of sodium sulphate contains 12.046 × 10²³ atoms of sodium.
(c) The mass percentage of sodium, sulphur and oxygen are 45.07, 32.39 and 22.54% respectively.
(d) The molar mass of sodium sulphate is 142 g mol-1.
Answer: (c)
Hint:
Mass % of an element = \(\frac{\left(\begin{array}{c}
\text { mass of that element } \\
\text { in the compound }
\end{array}\right) \times 100}{\text { Molar mass of the compound }}\)
Mass % of Na = \(\frac{46}{142}\) × 100 = 32.39 %
Mass % of S = \(\frac{32}{142}\) × 100 = 22.54 %
Mass % of O = \(\frac{64}{142}\) × 100 = 45.07 %

TN Board 11th Chemistry Important Questions

Must Read:

• CROMPTON Pivot Calculator