20 mL of sodium iodide solution gives 4.74 g of silver iodide when treated with excess silver nitrate solution. The molarity of the sodium iodide solution is: (Molar mass of AgI = 235 g/mol)
A1 M
B2 M
C0.5 M
D3 M
✅ Correct: A
Moles of AgI \(= 4.74/235 = 0.02\) mol. Since NaI + AgNO₃ → AgI (1:1 ratio), moles of NaI \(= 0.02\) mol. Molarity \(= 0.02/0.020\text{ L} = \mathbf{1\text{ M}}\).
Q2
A 0.05 cm thick coating of silver is deposited on a plate of area 0.05 m². Number of silver atoms deposited is \(N \times 10^{23}\). Find N. (Atomic mass Ag = 108, density = 7.9 g/cm³)
A7
B9
C11
D13
✅ Correct: C
Volume of coating \(= 0.05\text{ cm}\times(0.05\times10^4\text{ cm}^2) = 0.05\times500 = 25\text{ cm}^3\). Mass \(= 25\times7.9 = 197.5\text{ g}\). Moles \(= 197.5/108 \approx 1.83\) mol. Atoms \(= 1.83\times6.02\times10^{23} \approx \mathbf{11\times10^{23}}\). So N \(= 11\).
Q3
The mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess dilute HCl is: (Mg + 2HCl → MgCl₂ + H₂; molar volume at STP = 22400 mL)
A0.24 mg
B235.7 g
C2.444 g
D236 mg
✅ Correct: D
Moles of H₂ \(= 220/22400 = 9.82\times10^{-3}\) mol. Moles of Mg required \(= 9.82\times10^{-3}\) mol (1:1 ratio). Mass of Mg \(= 9.82\times10^{-3}\times24 = 0.2357\text{ g} = \mathbf{236\text{ mg}}\).
Q4
Molality of an aqueous solution of urea is 4.44 m. The mole fraction of urea in the solution is \(x \times 10^{-2}\). Find x. (Molar mass of urea = 60 g/mol, water = 18 g/mol)
A5
B7.4
C8
D3.6
✅ Correct: B
4.44 m means 4.44 mol urea in 1000 g water. Moles of water \(= 1000/18 = 55.56\) mol. Mole fraction of urea \(= 4.44/(4.44+55.56) = 4.44/60 = 0.074 = 7.4\times10^{-2}\). So x \(= \mathbf{7.4}\).
Q5
X g of benzoic acid (C₆H₅COOH, M = 122 g/mol) on reaction with aqueous NaHCO₃ released CO₂ gas which occupied 11.2 L at STP. The value of X is:
A30.5 g
B61 g
C122 g
D244 g
✅ Correct: B
Moles of CO₂ \(= 11.2/22.4 = 0.5\) mol. From reaction: C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂ (1:1 ratio). Moles of benzoic acid \(= 0.5\) mol. Mass \(= 0.5\times122 = \mathbf{61\text{ g}}\).
Q6
The molality (m) of a 3 M aqueous solution of NaCl is ____. (Density of solution = 1.25 g/mL, molar mass NaCl = 58.5 g/mol)
A2.90 m
B3.85 m
C2.79 m
D1.90 m
✅ Correct: C
1 L solution has mass \(= 1000\times1.25 = 1250\text{ g}\). Moles NaCl \(= 3\). Mass of NaCl \(= 3\times58.5 = 175.5\text{ g}\). Mass of water \(= 1250-175.5 = 1074.5\text{ g} = 1.0745\text{ kg}\). Molality \(= 3/1.0745 = \mathbf{2.79\text{ m}}\).
Q7
The empirical formula of a compound is CH₂O. Its molar mass is 90 g/mol. The molecular formula is:
AC₂H₄O₂
BC₃H₆O₃
CCH₂O
DC₄H₈O₄
✅ Correct: B
Empirical formula mass of CH₂O \(= 12+2+16 = 30\text{ g/mol}\). \(n = 90/30 = 3\). Molecular formula \(= 3\times\text{CH}_2\text{O} = \mathbf{C_3H_6O_3}\).
Q8
A solution is made by dissolving 1.5 mol NaOH in 500 g of water. Molality of NaOH in the solution is:
A1.5 m
B2.0 m
C3.0 m
D0.75 m
✅ Correct: C
Molality \(= \dfrac{\text{moles of solute}}{\text{kg of solvent}} = \dfrac{1.5}{0.500} = \mathbf{3.0\text{ m}}\).
Q9
In the reaction: 2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 8H₂O + 5Cl₂. The equivalent weight of KMnO₄ in acidic medium (molar mass = 158 g/mol) is:
A158
B79
C31.6
D52.7
✅ Correct: C
In acidic medium, Mn goes from +7 to +2 → change of 5 electrons per KMnO₄. n-factor \(= 5\). Equivalent weight \(= 158/5 = \mathbf{31.6\text{ g/eq}}\).
Q10
How many moles of electrons are transferred when 96500 C of electricity is passed through a solution? (F = 96500 C/mol)
A1 mol
B2 mol
C0.5 mol
D96500 mol
✅ Correct: A
1 Faraday \(= 96500\text{ C} = 1\) mol of electrons. When 96500 C is passed, \(\mathbf{1\text{ mol}}\) of electrons is transferred. This is the definition of the Faraday constant.
Q11
28 g of N₂ and 32 g of O₂ are mixed. The mole fraction of N₂ in the mixture is:
A0.25
B0.5
C0.75
D1
✅ Correct: B
Moles of N₂ \(= 28/28 = 1\) mol. Moles of O₂ \(= 32/32 = 1\) mol. Total \(= 2\) mol. Mole fraction of N₂ \(= 1/2 = \mathbf{0.5}\).
Q12
The number of atoms present in 4.25 g of NH₃ is: (N = 14, H = 1; Avogadro's number = 6.02 × 10²³)
A\(6.02\times10^{23}\)
B\(1.505\times10^{23}\)
C\(6.02\times10^{24}\)
D\(6.02\times10^{22}\)
✅ Correct: A
Moles of NH₃ \(= 4.25/17 = 0.25\) mol. Molecules \(= 0.25\times6.02\times10^{23} = 1.505\times10^{23}\). Each NH₃ has 4 atoms (1N + 3H). Total atoms \(= 4\times1.505\times10^{23} = \mathbf{6.02\times10^{23}}\).
Q13
1 mol of an ideal gas at STP occupies 22.4 L. What volume will 0.1 mol of CO₂ occupy at STP?
A2.24 L
B22.4 L
C224 L
D0.224 L
✅ Correct: A
Volume \(= n\times22.4 = 0.1\times22.4 = \mathbf{2.24\text{ L}}\). At STP, molar volume = 22.4 L/mol for all ideal gases.
Q14
The mass of one molecule of water is:
A\(3.0\times10^{-23}\text{ g}\)
B\(2.99\times10^{-23}\text{ g}\)
C\(18\text{ g}\)
D\(6.02\times10^{23}\text{ g}\)
✅ Correct: A
Molar mass of H₂O \(= 18\text{ g/mol}\). Mass of one molecule \(= 18/(6.02\times10^{23}) = \mathbf{3.0\times10^{-23}\text{ g}}\).
Q15
A compound contains 40% C, 6.67% H, and 53.33% O by mass. Its empirical formula is: (C=12, H=1, O=16)
ACH₂O
BC₂H₄O₂
CCHO
DCH₂O₂
✅ Correct: A
Mole ratio: C \(= 40/12 = 3.33\), H \(= 6.67/1 = 6.67\), O \(= 53.33/16 = 3.33\). Dividing by 3.33: C:H:O = 1:2:1. Empirical formula \(= \mathbf{CH_2O}\).
Q16
The normality of a solution containing 4.9 g of H₂SO₄ in 500 mL of solution is: (Molar mass H₂SO₄ = 98 g/mol, n-factor = 2)
The number of moles of electrons gained by 1 mol of Cr₂O₇²⁻ when it acts as oxidising agent in acidic medium (Cr⁶⁺ → Cr³⁺) is:
A2
B3
C6
D12
✅ Correct: C
Each Cr⁶⁺ gains 3 electrons to become Cr³⁺. There are 2 Cr atoms in Cr₂O₇²⁻. Total electrons gained \(= 2\times3 = \mathbf{6\text{ mol}}\) per mole of Cr₂O₇²⁻. n-factor of K₂Cr₂O₇ in acidic medium = 6.
Q19
Which of the following has the highest number of atoms?
A1 g of H₂
B1 g of O₂
C1 g of N₂
D1 g of CH₄
✅ Correct: A
Atoms = (mass/molar mass) × Avogadro number × atoms per molecule. H₂: \((1/2)\times6.02\times10^{23}\times2 = 6.02\times10^{23}\). O₂: \((1/32)\times2 = 3.76\times10^{22}\). N₂: \((1/28)\times2 = 4.3\times10^{22}\). CH₄: \((1/16)\times5 = 1.88\times10^{23}\). 1 g H₂ has most atoms.
Q20
Molarity of pure water (density = 1 g/mL, molar mass = 18 g/mol) is:
A18 M
B1 M
C55.5 M
D0.555 M
✅ Correct: C
1 L of water has mass = 1000 g. Moles of water \(= 1000/18 = 55.5\) mol. In 1 L: Molarity \(= \mathbf{55.5\text{ M}}\). This is an important concept — pure water is ~55.5 M.
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