CBSE Class 12 Chemistry Solutions NEET JEE Questions Set 01

Single Correct Answer

Question 1. Which of the following is dependent on temperature? (NEET 2017)
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Weight percentage
Answer:
Ans. (b)
Molarity of a solution depends upon the volume of the solution which changes with temperature. All others depend only on the masses of the solute and solvent which do not change with temperature.

Teacher's Note:
1. Any concentration term with volume in it (molarity, normality, mass/volume %) changes with temperature.

Question 2. A 5.2 molal aqueous solution of methyl alcohol, CH₃OH, is supplied. What is the mole fraction of methyl alcohol in the solution? (AIEEE 2011)
(a) 0.190
(b) 0.086
(c) 0.050
(d) 0.100
Answer:
Ans. (b)
5.2 molal solution means 5.2 moles of the solute (CH₃OH) are present in 1000 g of water, i.e., 1000/18 moles = 55.55 moles of water.
∴ Mole fraction of CH₃OH \( = \frac{5.2}{5.2 + 55.55} \) = 0.086

Teacher's Note:
1. From molality, take 1000 g of solvent: moles of water = 55.55, then use x = n₂/(n₁ + n₂).

Question 3. The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be (JEE Main 2013)
(a) 0.975 M
(b) 0.875 M
(c) 1.00 M
(d) 1.175 M
Answer:
Ans. (b)
M₁V₁ + M₂V₂ = M₃(V₁ + V₂)
0.5 × 750 + 2 × 250 = M₃(750 + 250)
375 + 500 = 1000 M₃ or M₃ \( = \frac{875}{1000} \) = 0.875 M

Teacher's Note:
1. On mixing solutions of the same solute, total moles add up and total volume adds up; the molarity is a weighted average.

Question 4. The density (in g mL⁻¹) of a 3.60 M sulphuric acid solution that is 29% H₂SO₄ (Molar mass = 98 g mol⁻¹) by mass will be (AIEEE 2007)
(a) 1.45
(b) 1.64
(c) 1.88
(d) 1.22
Answer:
Ans. (d)
29% H₂SO₄ by mass means 29 g H₂SO₄ are present in 100 g of the solution. If d g mL⁻¹ is the density of the solution, volume \( = \frac{100}{d} \) mL.
Further, 29 g H₂SO₄ \( = \frac{29}{98} \) mole. Hence, molarity of the solution \( = \frac{29/98}{100/d} \times 1000 \) mol L⁻¹ = 3.60 mol L⁻¹ (Given)
i.e., \( \frac{29}{98} \times \frac{1}{100} \times 1000 \times d = 3.60 \) or \( d = \frac{3.60}{10} \times \frac{98}{29} \) = 1.22 g mL⁻¹

Teacher's Note:
1. Use M = (10 × d × mass %)/molar mass to switch quickly between mass per cent, density and molarity.

Question 5. Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mL⁻¹. Volume of the acid required to make one litre of 0.1 M H₂SO₄ solution is
(a) 5.55 mL
(b) 11.10 mL
(c) 16.65 mL
(d) 22.20 mL
Answer:
Ans. (a)
98% H₂SO₄ by mass means 98 g H₂SO₄ present in 100 g of the solution, i.e., 100/1.80 mL = 55.5 mL
Also 98 g H₂SO₄ = 1 mole H₂SO₄. ∴ Molarity of 98% H₂SO₄ solution \( = \frac{1}{55.5} \times 1000 \) mol L⁻¹
Now, M₁V₁ = M₂V₂. Hence, \( \frac{1000}{55.5} \) M × V₁ = 1000 mL × 0.1 M or V₁ = 5.55 mL

Teacher's Note:
1. First find the molarity of the concentrated acid (here 18 M), then apply the dilution law M₁V₁ = M₂V₂.

Question 6. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated nitric acid is 70% HNO₃ (AIPMT 2013)
(a) 45.0 g conc HNO₃
(b) 90.0 g conc HNO₃
(c) 70.0 g conc HNO₃
(d) 54.0 g conc HNO₃
Answer:
Ans. (a)
250 mL of 2.0 M HNO₃ will contain HNO₃ \( = \frac{2}{1000} \times 250 \) = 0.5 mol = 0.5 × 63 g = 31.5 g
As concentrated HNO₃ is 70%, therefore, concentrated HNO₃ required \( = \frac{100}{70} \times 31.5 \) = 45 g

Teacher's Note:
1. The answer is the mass of the 70% solution, not of pure HNO₃; divide the pure mass by 0.70.

Question 7. 6.02 × 10²⁰ molecules of urea are present in 100 mL of its solution. The concentration of the solution is (AIPMT 2013)
(a) 0.02 M
(b) 0.01 M
(c) 0.001 M
(d) 0.1 M
Answer:
Ans. (b)
No. of moles of urea present in 100 mL of solution \( = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \) mol
∴ Molar concentration of urea in the solution \( = \frac{10^{-3}}{100} \times 1000 = 10^{-2} \) M = 0.01 M

Teacher's Note:
1. Convert molecules to moles with Avogadro's number first; remember to convert 100 mL to litres.

Question 8. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H₃PO₃), the volume of 0.1 M aqueous KOH solution required is (AIPMT 2013)
(a) 10 mL
(b) 20 mL
(c) 40 mL
(d) 60 mL
Answer:
Ans. (c)
H₃PO₃ is dibasic while KOH is monoacidic.
n_aM_aV_a = n_bM_bV_b, 2 × 0.1 × 20 = 1 × 0.1 × V_b or V_b = 40 mL.

Teacher's Note:
1. H₃PO₃ has only two ionisable P–OH hydrogens (one H is bonded directly to P), so its basicity is 2, not 3.

Question 9. The volumes of 4 N HCl and 10 N HCl required to make 1 litre of 6 N HCl are
(a) 0.75 litre of 4 N HCl and 0.25 litre of 10 N HCl
(b) 0.25 litre of 4 N HCl and 0.75 litre of 10 N HCl
(c) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl
(d) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl
(e) 0.50 litre of 4 N HCl and 0.50 litre of 10 N HCl
Answer:
Ans. (c)
Suppose x L of 4 N HCl are taken. Then 10 N HCl taken = (1 − x) L. Applying N₁V₁ + N₂V₂ = N₃V₃
4 × x + 10(1 − x) = 6 × 1 or 6x = 4 or \( x = \frac{4}{6} \) = 0.67 L.

Teacher's Note:
1. Since 6 N is closer to 4 N than to 10 N, more of the 4 N acid must be taken; this quickly rules out options (b) and (e).

Question 10. A person is considered to be suffering from lead poisoning if his concentration in him is more than 10 micrograms of lead per decilitre of blood. Concentration in parts per billion parts is
(a) 1
(b) 10
(c) 100
(d) 1000
Answer:
Ans. (c)
10 μg = 10 × 10⁻⁶ g = 10⁻⁵ g; 1 decilitre = 0.1 L = 100 mL
Thus, 100 mL of blood contain lead = 10⁻⁵ g
∴ 1 billion (10⁹) parts will contain lead \( = \frac{10^{-5}}{100} \times 10^{9} = 10^{2} \) = 100 parts
i.e., concentration = 100 parts per billion parts

Teacher's Note:
1. For dilute aqueous liquids take 1 mL ≈ 1 g; ppb = (mass of component/total mass) × 10⁹.

Question 11. The hardness of a water sample (in terms of equivalents of CaCO₃) containing 10⁻³ M CaSO₄ is: (molar mass of CaSO₄ = 136 g mol⁻¹) (JEE Main 2019)
(a) 100 ppm
(b) 50 ppm
(c) 10 ppm
(d) 90 ppm
Answer:
Ans. (a)
10⁻³ M means 10⁻³ moles of CaSO₄ present in 1 L of water.
Mass of 10⁻³ moles of CaSO₄ = 10⁻³ × 136 g = 0.136 g
Thus, 0.136 g of CaSO₄ are present in 1 L of water, i.e., 1 kg of water or 1000 g of water or 10³ g of water
⇒ 10⁶ g of water will contain CaSO₄ \( = \frac{0.136}{10^{3}} \times 10^{6} \) = 136 g
136 g CaSO₄ ≡ 100 g CaCO₃
∴ Hardness of water in terms of equivalents of CaCO₃ = 100 ppm.

Teacher's Note:
1. 1 mol of any Ca²⁺ salt is equivalent to 1 mol (100 g) of CaCO₃; so 10⁻³ M Ca²⁺ = 0.1 g CaCO₃ per litre = 100 ppm.

Question 12. Calculate the molarity of 3.3 molal solution of KCl whose density is 1.28 g mL⁻¹ (JEE Main 2021)
(a) 3.7 M
(b) 5.0 M
(c) 3.4 M
(d) 2.5 M
Answer:
Ans. (c)
3.3 mol are present in 1000 g of the solvent. 3.3 mol of KCl = 3.3 × 74.5 g = 245.8 g
∴ Mass of solution = 1000 + 245.8 g = 1245.8 g
Volume of solution \( = \frac{1245.8}{1.28} \) = 973.3 mL
Molarity \( = \frac{3.3}{973.3} \times 1000 \) = 3.39 ≈ 3.4 M

Teacher's Note:
1. Mass of solution = mass of solvent + mass of solute; forgetting the solute mass is the most common error here.

Question 13. 250 g of D-glucose solution in water contains 10.8% of carbon by weight. The molality of the solution is nearest to (Given: Atomic weights are H = 1 u, C = 12 u, O = 16 u) (JEE Main 2022)
(a) 1.03
(b) 2.06
(c) 3.09
(d) 5.40
Answer:
Ans. (b)
Glucose is C₆H₁₂O₆ (M = 180)
Mass of carbon in 250 g of solution \( = \frac{10.8}{100} \times 250 \) = 27 g
As 180 g of glucose contains 72 g of carbon (6 C-atoms), mass of glucose \( = \frac{180}{72} \times 27 \) = 67.5 g
∴ Mass of water = 250 − 67.5 = 182.5 g
∴ Molality of solution \( = \frac{67.5/180}{182.5} \times 1000 \) = 2.055 ≈ 2.06

Teacher's Note:
1. All the carbon comes from glucose, so the carbon percentage tells you how much glucose is present.

Question 14. In one molal solution that contains 0.5 mole of a solute, there is (NEET 2022)
(a) 500 mL of solvent
(b) 500 g of solvent
(c) 100 mL of solvent
(d) 1000 g of solvent
Answer:
Ans. (b)
Molality is the moles of the solute dissolved in 1000 g of the solvent.
∴ 1 molal solution means 1 mole of the solute are dissolved in 1000 g of the solvent
∴ 0.5 mole of the solute are dissolved in 500 g of the solvent.

Teacher's Note:
1. Molality is defined per kilogram of solvent (mass), so options in mL cannot be right.

Question 15. Which one of the following statements is not true?
(a) Dissolution of all solid solutes in water is exothermic
(b) Common salt is more soluble in water than canesugar at the same temperature
(c) Solubility of sodium sulphate decahydrate crystals first increases upto a certain temperature and then decreases
(d) Enthalpy of solution can be found using Clausius-Clapeyron equation
Answer:
Ans. (a)
Dissolution of some solids is exothermic while that of some others is endothermic.

Teacher's Note:
1. Solids like NH₄Cl, KNO₃ and urea dissolve endothermically (the solution cools), which is why their solubility rises with temperature.

Question 16. The mole fraction of a gas dissolved in a solvent is given by Henry's law. If the Henry's law constant for a gas in water at 298 K is 5.55 × 10⁷ Torr and the partial pressure of the gas is 200 Torr, then what is the amount of the gas dissolved in 1.0 kg of water?
(a) 2.0 × 10⁻⁴ mol
(b) 2.5 × 10⁻⁵ mol
(c) 3.7 × 10⁻⁶ mol
(d) 1.2 × 10⁻⁸ mol
Answer:
Ans. (a)
By Henry's law, p_A = K_H × x_A
or \( x_A = \frac{p_A}{K_H} = \frac{200\ \text{Torr}}{5.55 \times 10^{7}\ \text{Torr}} = 3.6 \times 10^{-6} \)
But \( x_A = \frac{n_A}{n_A + n_{H_2O}} \approx \frac{n_A}{n_{H_2O}} = \frac{n_A}{1000/18} \)
∴ \( n_A = x_A \times \frac{1000}{18} = 3.6 \times 10^{-6} \times \frac{1000}{18} \) mole = 2.0 × 10⁻⁴ mole.

Teacher's Note:
1. Because x_A is tiny, n_A can be neglected in the denominator: n_A ≈ x_A × n_water.

Question 17. The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is 4 × 10⁻³ kg L⁻¹. Therefore, the mass of the gas in kg dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K is
(a) 2.5 × 10⁻³
(b) 2.0 × 10⁻³
(c) 1.25 × 10⁻³
(d) 5.0 × 10⁻³
(e) 3 × 10⁻³
Answer:
Ans. (a)
By Henry's law, m = K_H p. At constant temperature, K_H is constant.
∴ At two different pressures, \( \frac{m}{m'} = \frac{p}{p'} \), \( \frac{4 \times 10^{-3}\ \text{kg L}^{-1}}{m'} = \frac{100}{250} \)
or \( m' = \frac{250}{100} \times 4 \times 10^{-3} \) kg L⁻¹ = 10⁻² kg L⁻¹ \( = \frac{10^{-2}}{4} \) kg in 250 mL = 2.5 × 10⁻³ kg.

Teacher's Note:
1. Two steps: scale the solubility with pressure (Henry's law), then scale with the volume asked (250 mL = L/4).

Question 18. For the solution of gases W, X, Y and Z in water at 298 K, the Henry's law constants (K_H) are 0.5, 2, 35 and 40 kbar respectively. The correct plot for the given data is (JEE Main 2019)
(a) Plot (a)
(b) Plot (b)
(c) Plot (c)
(d) Plot (d)

Answer:
Ans. (b)
According to Henry's law, p_gas = K_H x_gas ...(i)
or p_gas = K_H(1 − x_H2O) = K_H − K_H x_H2O
Hence, plot of p_gas vs x_H2O will be linear with negative slope. Further, for a fixed value of x_H2O, p_gas ∝ K_H. Hence, the correct plots are shown in option (b).

Teacher's Note:
1. Read the x-axis carefully: it is the mole fraction of water, not of the gas, so the lines slope downwards.

Question 19. The amount of solute (molar mass 60 g mol⁻¹) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is (Kerala PET 2010)
(a) 30 g
(b) 60 g
(c) 120 g
(d) 12 g
(e) 24 g
Answer:
Ans. (b)
If p° = 100 mm, p_s = 100 − 10 = 90 mm
\( \frac{p^\circ - p_s}{p^\circ} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2}{M_2} \times \frac{M_1}{w_1} \), \( \frac{100-90}{100} = \frac{w_2}{60} \times \frac{18}{180} \)
or \( w_2 = \frac{10}{100} \times 60 \times 10 \) = 60 g.

Teacher's Note:
1. A 10% lowering means relative lowering = 0.10; with dilute-solution approximation, n₂/n₁ = 0.10.

Question 20. At 80°C, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg. If a mixture solution of 'A' and 'B' boils at 80°C and 1 atm pressure, then amount of 'A' in the mixture is (1 atm = 760 mm Hg) (AIEEE 2008)
(a) 48 mol percent
(b) 50 mol percent
(c) 52 mol percent
(d) 34 mol percent
Answer:
Ans. (b)
P_total (at 80°C) = 760 mm
P_total = x_Ap_A° + x_Bp_B° = x_Ap_A° + (1 − x_A)p_B° = p_B° + x_A(p_A° − p_B°)
∴ 1000 + x_A(520 − 1000) = 760 or 480 x_A = 240 or x_A = 0.50, i.e., 50 mol percent.

Teacher's Note:
1. A liquid boils when its total vapour pressure equals the external pressure, so P_total = 760 mm here.

Question 21. Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1 : 1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1 : 2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are (Kerala PET 2008)
(a) 250 mm, 550 mm
(b) 350 mm, 450 mm
(c) 350 mm, 700 mm
(d) 500 mm, 500 mm
(e) 550 mm, 250 mm
Answer:
Ans. (e)
Molar ratio of 1 : 1 means x_X = x_Y = 0.5
P_total = x_Xp_X° + x_Yp_Y°. ∴ 400 = 0.5 p_X° + 0.5 p_Y° or p_X° + p_Y° = 800 ...(i)
Molar ratio of 1 : 2 means \( x_X = \frac{1}{3}, x_Y = \frac{2}{3} \). ∴ \( 350 = \frac{1}{3}p_X^\circ + \frac{2}{3}p_Y^\circ \) or p_X° + 2p_Y° = 1050 ...(ii)
Eqn. (ii) − Eqn. (i) gives p_Y° = 250 mm. Putting in (i), we get p_X° = 550 mm. Hence, p_X° = 550 mm, p_Y° = 250 mm.

Teacher's Note:
1. Mind the order asked (X first, then Y): options (a) and (e) contain the same numbers in opposite order.

Question 22. The relative lowering of vapour pressure of an aqueous solution containing non-volatile solute is 0.0125. The molality of the solution is
(a) 0.70
(b) 0.50
(c) 0.60
(d) 0.80
(e) 0.40
Answer:
Ans. (a)
Relative lowering of vapour pressure = mole fraction of the solute in the solution, i.e., \( \frac{n_2}{n_1 + n_2} = 0.0125 \)
Molality = n₂ when n₁ = 1000/18 mole = 55.55 mole
\( \frac{n_1 + n_2}{n_2} = \frac{1}{0.0125} = 80 \) or \( \frac{n_1}{n_2} + 1 = 80 \) or \( \frac{n_1}{n_2} = 79 \) or \( n_2 = \frac{n_1}{79} = \frac{55.55}{79} \) = 0.70

Teacher's Note:
1. Relative lowering equals the solute mole fraction; molality is then moles of solute per 55.55 mol (1 kg) of water.

One or More Correct Answers

Question 23. If P⁰ and P_S are the vapour pressures of the solvent and its solution respectively and N₁ and N₂ are the mole fractions of the solvent and solute respectively, then
(a) P_S = P⁰N₂
(b) P⁰ − P_S = P⁰N₂
(c) P_S = P⁰N₁
(d) (P⁰ − P_S)/P_S = N₁/(N₁ + N₂)
Answer:
Ans. (b, c)
By Raoult's law, P_S = P⁰N₁ (c is correct).
Then P⁰ − P_S = P⁰(1 − N₁) = P⁰N₂ (b is correct).
(a) is wrong as the vapour pressure is due to the solvent, and (d) is wrong because (P⁰ − P_S)/P_S = N₂/N₁.

Teacher's Note:
1. Relative lowering (P⁰ − P_S)/P⁰ equals the solute mole fraction N₂.

Passage-Based Questions

Question 24. Comprehension 1: An ideal solution of two liquids is a solution in which each component obeys Raoult's law which states that the vapour pressure of any component in the solution depends on the mole fraction of that component in the solution and the vapour pressure of that component in the pure state. However, there are many solutions which do not obey Raoult's law. In other words, they show deviations from ideal behaviour which may be positive or negative. However, in either case, corresponding to a particular composition, they form a constant boiling mixtures called azeotropes.
Which of the following mixture do you expect will not show positive deviation from Raoult's law?
(a) Benzene-Chloroform
(b) Benzene-Acetone
(c) Benzene-Ethanol
(d) Benzene-Carbon tetrachloride
Answer:
Ans. (a)
Benzene-Chloroform system shows negative deviations from Raoult's law.

Teacher's Note:
1. Chloroform's C–H can interact with the π-electrons of benzene, so the mixture has stronger A–B attraction, giving negative deviation.

Matrix Match Questions

Teacher's Note:
1. Convert everything to milliequivalents (meq = N × mL); for H₂SO₄, N = 2 × M.

Integer and Numerical Answer Questions

Question 26. Number of moles of Na₂CO₃ that should be dissolved in 4 litres of the solution to obtain 1 N Na₂CO₃ solution is
Answer:
Ans. 2
No. of moles of Na₂CO₃ required for 1 L of 1 N solution = 0.5
∴ No. of moles of Na₂CO₃ required for 4 L of 1 N solution = 4 × 0.5 = 2.

Teacher's Note:
1. Na₂CO₃ has n-factor 2, so 1 N = 0.5 M.

Question 27. The molality of a sulphuric acid solution in which the mole fraction of water is 0.86 is
Answer:
Ans. 9
Mole fraction of H₂SO₄ in the solution = 1 − 0.86 = 0.14
\( \frac{n_2}{n_1 + n_2} = 0.14 \). Molality is n₂ in 1000 g of water, i.e., in n₁ = 1000/18 = 55.55 moles
∴ \( \frac{n_2}{55.55 + n_2} = 0.14 \) or n₂ = 0.14n₂ + 7.777 or 0.86n₂ = 7.777 or n₂ = 9.0

Teacher's Note:
1. Shortcut: m = 1000 x₂/(x₁ M₁) = 1000 × 0.14/(0.86 × 18) = 9.0.

Question 28. 100 mL of 1 M H₂SO₄ are mixed with 200 mL of 8 M HCl solution. The normality of the resulting solution is
Answer:
Ans. 6
100 mL of 1 M H₂SO₄ = 100 millimoles = 200 milli eq
200 mL of 8 M HCl = 1600 millimoles = 1600 milli eq
Total milli eq = 200 + 1600 = 1800; Total volume = 100 + 200 mL = 300 mL
∴ Normality of resulting solution \( = \frac{1800}{300} \) = 6 N.

Teacher's Note:
1. Add milliequivalents (not millimoles) when finding normality of an acid mixture.

Question 29. The elevation in boiling point expected for 0.3 m Al₂(SO₄)₃ solution will be how many times compared with the elevation in boiling point of 0.1 m Na₂SO₄ solution?
Answer:
Ans. 5
ΔT_b = iK_b × m
For 0.3 m Al₂(SO₄)₃, ΔT_b = 5 × K_b × 0.3 = 1.5 K_b
For 0.1 m Na₂SO₄, ΔT_b = 3 × K_b × 0.1 = 0.3 K_b
Thus, ΔT_b for Al₂(SO₄)₃ solution will be 5 times that for Na₂SO₄ solution.

Teacher's Note:
1. Al₂(SO₄)₃ gives 2 + 3 = 5 ions; Na₂SO₄ gives 3 ions.

Question 30. If the compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is .......... molal (K_b = 0.52 K kg mol⁻¹) (JEE Main 2021)
Answer:
Ans. 3
AB → A⁺ + B⁻; 1 − α, α, α; i = 1 − α + α + α = 1 + α; α = 75% = 0.75 ∴ i = 1.75
ΔT_b = iK_bm; 2.5 = 1.75 × 0.52 × m or m = 2.75 mol kg⁻¹ ≈ 3 mol kg⁻¹ (nearest integer)

Teacher's Note:
1. For an AB-type electrolyte, i = 1 + α; divide ΔT_b by iK_b to get molality.

Assertion–Reason Questions

Question 31. Statement-1: If on mixing the two liquids, the solution becomes hot, it implies that it shows negative deviation from Raoult's law.
Statement-2: Solutions which show negative deviation are accompanied by decrease in volume.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of 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:
Ans. (b)
Correct explanation. In case of solutions showing negative deviation, the forces of interaction between the components are stronger than those in the pure components.

Teacher's Note:
1. Heat release (ΔH_mix < 0) and volume contraction (ΔV_mix < 0) are both consequences of stronger A–B attraction; one does not explain the other.

Question 32. Statement-1: If a liquid more volatile than the solvent is added to the solvent, the vapour pressure of the solution may increase, i.e., p_s > p°.
Statement-2: In the presence of more volatile liquid solute, Raoult's law does not hold good.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of 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:
Ans. (c)
Correct statement-2. Raoult's law is applicable to volatile solutes also.

Teacher's Note:
1. For a volatile solute, apply Raoult's law to both components: P = x₁p₁° + x₂p₂°, which can exceed p₁°.

Question 33. Statement-1: One molar aqueous solution has always higher concentration than one molal solution.
Statement-2: One molar solution contains less solvent than one molal solution.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of 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:
Ans. (a)
Statement-2 is the correct explanation of Statement-1.

Teacher's Note:
1. 1 M has 1 mol in 1 L of solution (less than 1 kg of water), while 1 m has 1 mol in a full 1 kg of water.

Question 34. Statement-1: Vapour pressure of water is less than 1.013 bar at 373 K.
Statement-2: Water boils at 373 K as the vapour pressure at this temperature becomes equal to atmospheric pressure.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation of 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:
Ans. (d)
Correct Statement-1. At 373 K, i.e., 100°C, the vapour pressure of water is equal to 1 atmosphere which is equal to 1.013 bar.

Teacher's Note:
1. Boiling point is the temperature at which vapour pressure equals the external pressure (1 atm = 1.013 bar).