CBSE Class 12 Chemistry Solutions NEET JEE Questions Set 02

Single Correct Answer

Question 1. If x₁ and x₂ represent the mole fraction of a component A in the vapour phase and liquid mixture respectively and p_A° and p_B° represent vapours pressures of pure A and pure B, then total vapour pressure of the liquid mixture is
(a) \( \frac{p_A^\circ x_1}{x_2} \)
(b) \( \frac{p_A^\circ x_2}{x_1} \)
(c) \( \frac{p_B^\circ x_1}{x_2} \)
(d) \( \frac{p_B^\circ x_2}{x_1} \)
Answer:
Ans. (b)
Mole fraction in the vapour phase \( (x_1) = \frac{p_A}{P_{total}} \)
But p_A = x_A × p_A° = x₂ × p_A°
Hence, \( x_1 = \frac{x_2 p_A^\circ}{P_{total}} \) or \( P_{total} = \frac{p_A^\circ x_2}{x_1} \)

Teacher's Note:
1. Combine Dalton's law (y_A = p_A/P) with Raoult's law (p_A = x_Ap_A°) for any vapour-composition question.

Question 2. An ideal solution is formed by mixing two volatile liquids A and B. X_A and X_B are the mole fractions of A and B respectively in the solution and Y_A and Y_B are the mole fractions of A and B respectively in the vapour phase. A plot of 1/Y_A along y-axis against 1/X_A along x-axis gives a straight line. What is the slope of the straight line? (where p_A° and p_B° are the vapour pressures of the pure components A and B respectively) (IAS Prelim 2010)
(a) p_B°/p_A°
(b) p_A°/p_B°
(c) p_B° − p_A°
(d) p_A° − p_B°
Answer:
Ans. (a)
Y_A = X_Ap_A°/P and P = p_B° + X_A(p_A° − p_B°), which gives
\( \frac{1}{Y_A} = \frac{p_B^\circ}{p_A^\circ}\cdot\frac{1}{X_A} + \frac{p_A^\circ - p_B^\circ}{p_A^\circ} \) (y = mx + c)
Hence, plot of 1/Y_A along y-axis versus 1/X_A along x-axis will be straight line with slope = p_B°/p_A°.

Teacher's Note:
1. This relation is worth remembering: 1/Y_A vs 1/X_A is linear with slope p_B°/p_A° and intercept 1 − p_B°/p_A°.

Question 3. The vapour pressure of a solvent decreases by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in vapour pressure is to be 20 mm of mercury?
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
Answer:
Ans. (b)
Δp/p° = x₂. Hence, Δp/Δp′ = x₂/x₂′, i.e., 10/20 = 0.2/x₂′ or x₂′ = 0.4. Hence, x₁′ = 1 − 0.4 = 0.6.

Teacher's Note:
1. The question asks for the solvent's mole fraction; 0.4 is the solute's, a common trap.

Question 4. Which of the following statements about the composition of the vapour over an ideal 1 : 1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C (Given vapour pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa) (NEET Phase I 2016)
(a) The vapour will contain equal amounts of benzene and toluene
(b) Not enough information is given to make a prediction
(c) The vapour will contain a higher percentage of benzene
(d) The vapour will contain a higher percentage of toluene
Answer:
Ans. (c)
P_Benzene = x_Benzene × P°_Benzene; P_Toluene = x_Toluene × P°_Toluene
For an ideal 1 : 1 mixture of benzene and toluene, \( x_{Benzene} = \frac{1}{2},\ x_{Toluene} = \frac{1}{2} \)
\( P_{Benzene} = \frac{1}{2} \times 12.8 \) kPa = 6.4 kPa; \( P_{Toluene} = \frac{1}{2} \times 3.85 \) kPa = 1.925 kPa
Thus, the vapour will contain higher percentage of benzene as the partial vapour pressure of benzene is higher than that of toluene.

Teacher's Note:
1. The vapour is always richer in the more volatile component (higher p°).

Question 5. Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time: (JEE Main 2020)
(a) the volume of the solution and the solvent does not change
(b) the volume of the solution increases and the volume of the solvent decreases
(c) the volume of the solution decreases and the volume of the solvent increases
(d) the volume of the solution does not change and the volume of the solvent decreases.
Answer:
Ans. (b)
Vapour pressure of solvent is higher than that of the solution. To attain equilibrium over time, vapour pressure of solvent beaker will decrease and that of the solution beaker will increase, i.e., some vapour from solvent beaker will condense into the solution beaker. Thus, volume of solution will increase and that of solvent will decrease.

Teacher's Note:
1. Solvent always moves from higher vapour pressure (pure solvent) to lower vapour pressure (solution), much like osmosis through the vapour.

Question 6. For an ideal solution, the correct option is (NEET 2019)
(a) Δ_mixG = 0 at constant T and P
(b) Δ_mixS = 0 at constant T and P
(c) Δ_mixV ≠ 0 at constant T and P
(d) Δ_mixH = 0 at constant T and P
Answer:
Ans. (d)
For an ideal solution, Δ_mixH = 0 at constant T and P.

Teacher's Note:
1. For ideal mixing ΔH = 0 and ΔV = 0, but ΔS > 0 and ΔG < 0, because mixing is spontaneous.

Question 7. Dry air is passed through a solution containing 10 g of the solute in 90 g of water and then through pure water. The loss in weight of solution is 2.5 g and that of pure solvent is 0.05 g. Calculate the molecular weight of the solute.
(a) 50
(b) 180
(c) 100
(d) 25
(e) 51
Answer:
Ans. (c)
Loss in weight of solution ∝ V.P. of solution (p_s); Loss in weight of solvent ∝ p° − p_s
∴ \( \frac{p^\circ - p_s}{p_s} = \frac{\text{Loss in weight of solvent}}{\text{Loss in weight of solution}} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2 M_1}{w_1 M_2} \)
\( \frac{0.05}{2.5} = \frac{10/M_2}{90/18} = \frac{2}{M_2} \) or M₂ = 100

Teacher's Note:
1. In the Ostwald–Walker method, loss from the solution bulbs ∝ p_s and loss from the solvent bulbs ∝ (p° − p_s).

Question 8. The vapour pressure of a solution of a non-volatile electrolyte (A) in solvent (B) is 95% of the vapour pressure of the solvent at the same temperature. If molar mass of B is 30% of molar mass of A, the mass ratio of the solvent and solute are
(a) 0.15
(b) 0.20
(c) 4.0
(d) 5.7
Answer:
Ans. (d)
p_s = 95% of p°, i.e., p_s = 0.95 p°; M_B = 30% of M_A, i.e., M_B = 0.30 M_A (B = solvent, A = electrolyte)
Applying complete formula, \( \frac{p^\circ - p_s}{p_s} = \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_A}{w_B} \times \frac{M_B}{M_A} \)
\( \frac{p^\circ - 0.95p^\circ}{0.95p^\circ} = \frac{w_A}{w_B} \times 0.30 \) or \( \frac{0.05}{0.95} = \frac{w_A}{w_B} \times 0.30 \)
or \( \frac{w_B}{w_A} = \frac{0.30 \times 0.95}{0.05} \) = 5.7

Teacher's Note:
1. The ratio asked is solvent : solute (w_B/w_A); inverting it gives 0.175, which is not an option, so check the order before choosing.

Question 9. If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g mol⁻¹) of the solute is [Given that molar mass of n-octane is 114 g mol⁻¹] (NEET 2020 Phase-1)
(a) 40
(b) 60
(c) 80
(d) 20
Answer:
Ans. (a)
\( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1} = \frac{w_2/M_2}{w_1/M_1} = \frac{w_2 M_1}{w_1 M_2} \)
\( \frac{100-80}{100} = \frac{8 \times 114}{114 \times M_2} \) or \( M_2 = 8 \times \frac{100}{20} \) = 40

Teacher's Note:
1. 114 g of octane is exactly 1 mol, so the relative lowering 0.2 directly gives n₂ = 0.2 mol.

Question 10. The empirical formula of a non-electrolyte is CH₂O. A solution containing 3 g L⁻¹ of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecular formula of the compound is (Karnataka CET 2011)
(a) CH₂O
(b) C₂H₄O₂
(c) C₄H₈O₄
(d) C₃H₆O₃
Answer:
Ans. (b)
Conc. of compound in solution = 3 g L⁻¹ \( = \frac{3}{M} \) mol L⁻¹
As it is isotonic with 0.05 M glucose solution, \( \frac{3}{M} = 0.05 \) or M = 60
Empirical formula mass of CH₂O = 30. ∴ \( n = \frac{\text{Mol. mass}}{\text{E.F. mass}} = \frac{60}{30} = 2 \)
Hence, molecular formula = 2 × CH₂O = C₂H₄O₂

Teacher's Note:
1. Isotonic solutions of non-electrolytes have equal molar concentrations at the same temperature.

Question 11. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol⁻¹) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm⁻³, molar mass of the substance will be (AIEEE 2007)
(a) 210.0 g mol⁻¹
(b) 90.0 g mol⁻¹
(c) 115.0 g mol⁻¹
(d) 105.0 g mol⁻¹
Answer:
Ans. (a)
Isotonic solutions have the same molar concentration. Hence, Molar conc. of the substance = Molar conc. of urea
\( \frac{52.5\ \text{g L}^{-1}}{M} = \frac{15\ \text{g L}^{-1}}{60\ \text{g mol}^{-1}} \) (100 g solution = 100 mL as d = 1 g mL⁻¹)
∴ \( M = \frac{52.5 \times 60}{15} \) = 210 g mol⁻¹.

Teacher's Note:
1. With density 1 g mL⁻¹, x% (w/w) means 10x g per litre.

Question 12. Insulin (C₂H₁₀O₅)_n is dissolved in a suitable solvent and the osmotic pressure (π) of solutions of various concentrations (C) is measured at 20°C. The slope of the plot of π (atm) versus C (in g/cm³) is found to be 4.65 × 10⁻³. The molecular weight of insulin is
(a) 3.17 × 10⁶
(b) 4.17 × 10⁶
(c) 5.17 × 10⁶
(d) 6.17 × 10⁶
Answer:
Ans. (c)
π = C′RT. Here, concentration C′ is in mol L⁻¹. To convert g cm⁻³ into mol L⁻¹, we have
C (g cm⁻³) \( = \frac{C}{M_2} \times 1000 \) mol L⁻¹
Thus, C′ in the formula should be replaced by this factor, i.e., \( \pi = \frac{C \times 1000}{M_2}RT = \frac{1000RT}{M_2}C \)
Thus, a plot of π vs C will be linear with slope \( = \frac{1000RT}{M_2} = 4.65 \times 10^{-3} \) (Given)
∴ \( \frac{1000 \times 0.0821 \times 293}{M_2} = 4.65 \times 10^{-3} \) or M₂ = 5.17 × 10⁶

Teacher's Note:
1. Watch the units of the concentration axis: g cm⁻³ must be converted to mol L⁻¹ (factor 1000/M) before using π = CRT.

Question 13. Osmotic pressure of insulin solution at 298 K is found to be 0.0072 atm. Hence, height of the water column due to this pressure will be (Given density of Hg = 13.6 g mL⁻¹)
(a) 7.4 mL
(b) 7.4 cm
(c) 74 cm
(d) 760 mm
Answer:
Ans. (b)
P = hdg. But P = 0.0072 atm. Hence, h = 0.0072 × 76 cm of Hg column, d = density of Hg = 13.6 g cm⁻³, g = 981 cm s⁻²
Hence, P = 0.0072 × 76 × 13.6 × 981 (for Hg column)
For water, d = 1 g cm⁻³, hence for water column P = h × 1 × 981
Thus, h × 981 = 0.0072 × 76 × 13.6 × 981 or h = 7.4 cm

Teacher's Note:
1. Heights of two liquid columns exerting the same pressure are inversely proportional to their densities: h_water = h_Hg × 13.6.

Question 14. A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in 125 cm³ of an aqueous solution. At 4°C, an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is (Assume density of solution is 1.00 g/cm³)
(a) 9.4 × 10⁵
(b) 5.4 × 10⁵
(c) 5.4 × 10¹⁰
(d) 9.4 × 10¹⁰
Answer:
Ans. (b)
Given h = 2.6 mm = 0.26 cm, V = 125 cm³. ∴ π = hdg = 0.26 × 1 × 980 dyne/cm²
\( M = \frac{wRT}{\pi V} = \frac{0.75 \times (8.314 \times 10^{7}) \times 277}{(0.26 \times 980)(125)} \) = 5.4 × 10⁵.

Teacher's Note:
1. In CGS units use R = 8.314 × 10⁷ erg K⁻¹ mol⁻¹ with π in dyne cm⁻² and V in cm³.

Question 15. The following solutions were prepared by dissolving
10 g of glucose (C₆H₁₂O₆) in 250 mL of water (P₁),
10 g of urea (CH₄N₂O) in 250 mL of water (P₂),
10 g of sucrose (C₁₂H₂₂O₁₁) in 250 mL of water (P₃).
The right option for the decreasing order of osmotic pressure of these solutions is: (NEET 2021)
(a) P₂ > P₁ > P₃
(b) P₁ > P₂ > P₃
(c) P₂ > P₃ > P₁
(d) P₃ > P₁ > P₂
Answer:
Ans. (a)
Osmotic pressure (π) = iCRT where C is molar concentration of the solution and i is van't Hoff factor. As all the compounds are non-electrolytes, i = 1 (same) for all. Thus, greater the C, greater is π.
As volume of solution in each case is same and mass dissolved is same, greater the molar mass, less is the number of moles and hence less is the molar concentration (C). Molar masses are glucose = 180, urea = 60, sucrose = 342.
Hence, order of molar concentration will be urea > glucose > sucrose and, therefore, order of π will be urea (P₂) > glucose (P₁) > sucrose (P₃).

Teacher's Note:
1. For equal masses of non-electrolytes in equal volumes, the colligative effect is inversely proportional to molar mass.

Question 16. An aqueous solution of urea is found to boil at 100.52°C. Given K_b for water is 0.52 K kg mol⁻¹, the mole fraction of urea in the solution is
(a) 1
(b) 0.5
(c) 0.018
(d) 0.25
Answer:
Ans. (c)
ΔT_b = K_b × m. Hence, molality, \( m = \frac{\Delta T_b}{K_b} = \frac{0.52}{0.52} = 1 \)
Molality = 1 means 1 mole of solute in 1000 g of solvent. But 1000 g of solvent (water) = 1000/18 moles = 55.55 moles
∴ Mole fraction of urea \( = \frac{1}{1 + 55.55} \) = 0.018.

Teacher's Note:
1. A 1 molal aqueous solution always has solute mole fraction 1/56.55 ≈ 0.018; a handy number to remember.

Question 17. For a dilute solution containing 2.5 g of a non-volatile, non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of the solute is much lower than the concentration of the solvent, the vapour pressure (mm of Hg) of the solution is (take K_b = 0.76 K kg mol⁻¹) (IIT 2012)
(a) 724
(b) 740
(c) 736
(d) 718
Answer:
Ans. (a)
\( M_2 = \frac{1000 K_b w_2}{w_1 \times \Delta T_b} = \frac{1000 \times 0.76 \times 2.5}{100 \times 2} \) = 9.5
\( \frac{p^\circ - p_s}{p^\circ} = x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} \) (as n₂ << n₁) \( = \frac{w_2/M_2}{w_1/M_1} \)
\( \frac{760 - p_s}{760} = \frac{2.5/9.5}{100/18} = 0.047 \) or 760 − p_s = 760 × 0.047 = 35.7 or p_s = 724.3 mm

Teacher's Note:
1. The solution boils at its normal boiling point when its vapour pressure is 1 atm; here p° of water at that temperature is taken as 760 mm.

Question 18. A solution containing 1.8 g of a compound (empirical formula CH₂O) in 40 g of water is observed to freeze at −0.465°C. The molecular formula of the compound is (K_f of water = 1.86 kg K mol⁻¹) (Kerala CET 2011)
(a) C₂H₄O₂
(b) C₃H₆
(c) C₄H₈O₄
(d) C₅H₁₀O₅
(e) C₆H₁₂O₆
Answer:
Ans. (e)
\( M_2 = \frac{1000 K_f w_2}{w_1 \times \Delta T_f} = \frac{1000 \times 1.86 \times 1.8}{40 \times 0.465} \) = 180
E.F. mass CH₂O = 30. ∴ \( n = \frac{180}{30} = 6 \)
∴ Molecular formula = 6 × CH₂O = C₆H₁₂O₆

Teacher's Note:
1. Find the molar mass from the colligative property first, then divide by the empirical formula mass.

Question 19. A solution containing 0.10 g of non-volatile solute X (molar mass: 100) in 200 g of benzene depresses the freezing point of benzene by 0.25°C while 0.50 g of another non-volatile solute Y in 100 g of benzene also depresses the freezing point of benzene by 0.25°C. What is the molecular mass of Y? (IAS Prelim 2010)
(a) 50
(b) 100
(c) 150
(d) 1000
Answer:
Ans. (d)
\( \Delta T_f = \frac{1000 K_f w_2}{w_1 \times M_2} = \frac{1000 K_f w_2'}{w_1' \times M_2'} \)
or \( M_2' = \frac{w_2'}{w_1'} \times \frac{w_1}{w_2} \times M_2 = \frac{0.50}{100} \times \frac{200}{0.10} \times 100 \) = 1000.

Teacher's Note:
1. Equal depression in the same solvent means equal molality; compare the two solutions directly without needing K_f.

Question 20. A solution of urea (mol. mass 56 g mol⁻¹) boils at 100.18°C at the atmospheric pressure. If K_f and K_b for water are 1.86 and 0.512 K kg mol⁻¹ respectively, the above solution will freeze at
(a) −6.54°C
(b) −0.654°C
(c) 6.54°C
(d) 0.654°C
Answer:
Ans. (b)
ΔT_b = K_b × m ∴ 0.18 = 0.512 × m or m = 0.18/0.512
\( \Delta T_f = K_f \times m = 1.86 \times \frac{0.18}{0.512} = 0.654 \)
∴ T_f = −0.654°C

Teacher's Note:
1. The same molality gives both effects, so ΔT_f/ΔT_b = K_f/K_b; the molar mass is not needed.

Question 21. At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If K_b = 0.52, the boiling point of this solution will be (NEET Phase I 2016)
(a) 102°C
(b) 103°C
(c) 101°C
(d) 100°C
Answer:
Ans. (c)
\( \frac{p^\circ - p_s}{p^\circ} = \frac{n_2}{n_1} = \frac{n_2}{w_1/M_1} \). ∴ \( \frac{760 - 732}{760} = \frac{n_2}{100/18} \)
or \( n_2 = \frac{28}{760} \times \frac{100}{18} \) = 0.2046 mole
Molality of the solution \( = \frac{0.2046\ \text{mole}}{100/1000\ \text{kg}} \) = 2.046 mole kg⁻¹
ΔT_b = K_b × molality = 0.52 × 2.046 = 1.06°, i.e., T_b − T_b° = 1.06°
or T_b = T_b° + 1.06 = 100 + 1.06°C = 101.06°C

Teacher's Note:
1. At 100°C, p° of water is 760 mm; get the moles of solute from relative lowering, then use ΔT_b = K_bm.

Question 22. In 100 g of naphthalene, 2.423 g of S was dissolved. Melting point of naphthalene = 80.1°C. ΔT_f = 0.661°C. L_f = 35.7 cal/g of naphthalene. Molecular formula of sulphur added is
(a) S₂
(b) S₄
(c) S₆
(d) S₈
Answer:
Ans. (d)
\( M_2 = \frac{1000 K_f w_2}{w_1 \Delta T_f} = \frac{1000 w_2}{w_1 \Delta T_f} \times \frac{RT_0^2}{1000 L_f} \)
\( = \frac{1000 \times 2.423}{100 \times 0.661} \times \frac{2 \times (353.1)^2}{1000 \times 35.7} \) = 256 g mol⁻¹
If S_x is the formula, then molecular mass x × 32 = 256 or x = 8. Hence, S₈.

Teacher's Note:
1. K_f = RT_f²/(1000 L_f) with R = 2 cal K⁻¹ mol⁻¹ when L_f is in cal g⁻¹; T must be in kelvin.

Question 23. K_f for water is 1.86 K kg mol⁻¹. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C₂H₆O₂) must you add to get the freezing point of the solution lowered to −2.8°C? (AIEEE 2012)
(a) 27 g
(b) 72 g
(c) 93 g
(d) 39 g
Answer:
Ans. (c)
\( M_2 = \frac{1000 K_f w_2}{w_1 \Delta T_f} \)
∴ \( 62 = \frac{1000 \times 1.86 \times w_2}{1000 \times 2.8} \) or w₂ = 93.3 g

Teacher's Note:
1. Molar mass of ethylene glycol is 62 g mol⁻¹; required molality = 2.8/1.86 = 1.5 mol kg⁻¹.

One or More Correct Answers

Question 24. The vapour pressure of a dilute solution of a solute is not influenced by
(a) nature of the solute if it is non-electrolyte
(b) mole fraction of the solute
(c) melting point of the solute
(d) degree of dissociation of the solute.
Answer:
Ans. (a, c)
Lowering of vapour pressure is a colligative property; it depends only on the number of solute particles, i.e., on the mole fraction of the solute and, for electrolytes, on the degree of dissociation.
It does not depend on the nature of a non-electrolyte solute or on its melting point. Hence, (a) and (c) are correct.

Teacher's Note:
1. Colligative properties count particles: anything that changes the number of particles (mole fraction, dissociation) matters; identity does not.

Question 25. In the depression of freezing point experiment, it is found that
(a) The vapour pressure of the solution is less than that of pure solvent
(b) The vapour pressure of the solution is more than that of pure solvent
(c) Only solute molecules solidify at the freezing point
(d) Only solvent molecules solidify at the freezing point.
Answer:
Ans. (a, d)
A non-volatile solute lowers the vapour pressure of the solvent, so the solution's vapour pressure equals that of the solid solvent at a lower temperature (a is correct).
At the freezing point of a dilute solution, only the solvent separates as a pure solid (d is correct).

Teacher's Note:
1. On freezing a dilute solution, pure solvent crystallises out and the remaining solution becomes more concentrated.

Passage-Based Questions

Question 26. 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.
An azeotropic solution of two liquids has boiling point lower than either of the two liquids when it
(a) shows no deviations from Raoult's law
(b) shows a positive deviation from Raoult's law
(c) shows a negative deviation from Raoult's law
(d) is saturated.
Answer:
Ans. (b)
Lower boiling point means higher vapour pressure than expected which implies positive deviation.

Teacher's Note:
1. Minimum-boiling azeotropes come from positive deviation (e.g., ethanol–water); maximum-boiling azeotropes from negative deviation (e.g., HNO₃–water).

Matrix Match Questions

Teacher's Note:
1. A break in a solubility curve marks a change in the solid phase in equilibrium with the solution (e.g., hydrate → anhydrous salt).

Teacher's Note:
1. i can be less than 1 (association) but never zero or negative.

Integer and Numerical Answer Questions

Question 29. 29.2% (w/w) HCl stock solution has a density of 1.25 g mL⁻¹. The molecular weight of HCl is 36.5 g mol⁻¹. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is (IIT 2012)
Answer:
Ans. 8
Stock solution of HCl = 29.2% (w/w). Thus, 29.2 g of HCl are present in 100 g of the solution
As density of the solution = 1.25 g mL⁻¹, volume of 100 g of the solution \( = \frac{100}{1.25} \) mL
As molar mass of HCl = 36.5 g mol⁻¹, molarity of the solution \( = \frac{29.2}{36.5} \times \frac{1.25}{100} \times 1000 \) = 10 M
M₁V₁ (stock solution) = M₂V₂ (solution required); 10 × V₁ = 0.4 × 200 or V₁ = 8 mL

Teacher's Note:
1. Molarity of a stock solution = (10 × % w/w × density)/molar mass.

Question 30. A compound H₂X with molar weight of 80 g is dissolved in a solvent having density of 0.4 g mL⁻¹. Assuming no change in volume upon dissolution, the molality of a 3.2 molar solution is (JEE Advanced 2014)
Answer:
Ans. 8
3.2 molar solution contains 3.2 moles of H₂X in 1000 mL of the solution. As there is no change in volume on dissolution, volume of solvent = 1000 mL.
Mass of solvent = 1000 mL × 0.4 g mL⁻¹ = 400 g = 0.4 kg
Molality \( = \frac{3.2\ \text{moles}}{0.4\ \text{kg}} \) = 8 mol kg⁻¹ = 8 m

Teacher's Note:
1. 'No change in volume' lets you take the solvent volume equal to the solution volume; then density gives the solvent mass.

Question 31. MX₂ dissociates into M²⁺ and X⁻ ions in an aqueous solution with a degree of dissociation (α) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is (JEE Advanced 2014)
Answer:
Ans. 2
MX₂ ⇄ M²⁺ + 2X⁻; 1 − α, α, 2α
i = 1 − α + α + 2α = 1 + 2α. But α = 0.5, ∴ i = 1 + 2 × 0.5 = 2
This is the required ratio, i.e., 2.

Teacher's Note:
1. The ratio of observed to normal colligative property is the van't Hoff factor itself.

Question 32. If the freezing point of a 0.01 molal aqueous solution of cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is −0.0558°C, the number of chloride(s) in the coordination sphere of the complex is (K_f of water = 1.86 K kg mol⁻¹) (JEE Advanced 2015)
Answer:
Ans. 1
ΔT_f = iK_fm. Given m = 0.01 molal, T_f = −0.0558°C, K_f = 1.86 K kg mol⁻¹
ΔT_f = T_f° − T_f = 0 − (−0.0558) = 0.0558°
\( i = \frac{\Delta T_f}{K_f m} = \frac{0.0558}{1.86 \times 0.01} = 3 \)
As 3 ions are produced on dissociation of the complex, the molecular formula of the complex is [Co(NH₃)₅Cl]Cl₂. Thus, only one chloride ion is present in the coordination sphere.

Teacher's Note:
1. Co(III) needs 3 Cl in total; i = 3 means 2 Cl⁻ are outside, so 3 − 2 = 1 Cl is inside the coordination sphere.

Question 33. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is same as its molality. Density of the solution at 298 K is 2.0 g cm⁻³. The ratio of the molecular weights of the solute and solvent \( \left(\frac{MW_{solute}}{MW_{solvent}}\right) \), is (JEE Advanced 2016)
Answer:
Ans. 9
Mole fraction of solute = 0.1 means moles of solute = 0.1, moles of solvent = 0.9
Mass of solute = 0.1 × M_solute g; Mass of solvent = 0.9 × M_solvent g
Total mass of solution = (0.1 M_solute + 0.9 M_solvent) g
Total volume of solution \( = \frac{\text{Mass}}{\text{Density}} = \frac{0.1M_{solute} + 0.9M_{solvent}}{2 \times 1000} \) L
Molarity of solution \( = \frac{0.1 \times 2000}{0.1M_{solute} + 0.9M_{solvent}} = \frac{200}{0.1M_{solute} + 0.9M_{solvent}} \) mol L⁻¹
Molality of solution \( = \frac{0.1 \times 1000}{0.9M_{solvent}} = \frac{100}{0.9M_{solvent}} \) mol kg⁻¹
As molarity = molality, \( \frac{200}{0.1M_{solute} + 0.9M_{solvent}} = \frac{100}{0.9M_{solvent}} \)
or 0.1 M_solute + 0.9 M_solvent = 2 × 0.9 M_solvent
or \( \frac{0.1M_{solute}}{M_{solvent}} + 0.9 = 1.8 \) or \( \frac{M_{solute}}{M_{solvent}} = \frac{0.9}{0.1} \) = 9

Teacher's Note:
1. Molarity = molality means 1 L of solution contains exactly 1 kg of solvent; here 2000 g of solution has 1000 g solvent and 1000 g solute.

Question 34. 1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is .......... (Nearest integer) (Given: Density of acetic acid is 1.02 g mL⁻¹, Molar mass of acetic acid is 60 g mol⁻¹, K_f(H₂O) = 1.85 K kg mol⁻¹) (JEE Main 2022)
Answer:
Ans. 5
ΔT_f = iK_fm. Moles of solute (acetic acid) \( = \frac{1.2 \times 1.02}{60} \)
As moles of solute are very less, molarity ≈ molality. \( 0.0198 = i \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2} \) or i = 1.05
CH₃COOH → CH₃COO⁻ + H⁺; 1 − α, α, α
∴ i = 1 + α or α = i − 1 = 1.05 − 1 = 0.05 or 5%

Teacher's Note:
1. Convert volume of liquid acid to mass with its density, then to moles; for a very dilute solution, molarity ≈ molality.

Assertion–Reason Questions

Question 35. Statement-1: Any concentration of NaCl solution can be injected intravenously as NaCl, being a common table salt, is a harmless chemical.
Statement-2: 0.9% (mass/volume) NaCl solution is isotonic with the fluid inside the blood cells.
(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. Only 0.9% NaCl solution can be injected intravenously because it is isotonic with the fluid inside the blood cells, otherwise they will shrink or swell.

Teacher's Note:
1. Hypertonic saline makes blood cells shrink (plasmolysis) and hypotonic solution makes them swell and burst (haemolysis).

Question 36. Assertion: One molar aqueous solution has always higher concentration than one molal.
Reason: The molality of a solution depends upon the density of the solution whereas molarity does not. (AIIMS 2010)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
One molar aqueous solution contains one mole of the solute in 1000 mL of the solution whereas one molal solution contains one mole of the solute in 1000 g of the solvent. Thus, solvent present in one molar solution is less than 1000 mL. Hence, concentration is more.
Correct R. The molarity of a solution depends upon the density (volume) of the solution whereas molality does not.

Teacher's Note:
1. Molality uses masses only, so it is independent of density and temperature; it is molarity that needs the density to convert.

Question 37. Assertion: Greater the value of Henry's constant of a gas in a particular solvent, greater is the solubility of the gas at the same pressure and temperature.
Reason: Solubility of a gas is directly proportional to its Henry's constant at the same pressure and temperature.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct A. Greater the value of Henry's constant of a gas in a particular solvent, less is the solubility of the gas at the same pressure and temperature.
Correct R. Solubility of a gas is inversely proportional to its Henry's constant at the same pressure and temperature.

Teacher's Note:
1. From p = K_Hx, x = p/K_H: a large K_H means a less soluble gas.

Question 38. Assertion: If a liquid solute more volatile than the solvent is added to a solvent, the vapour pressure of the solution may increase, i.e., p_s > p°.
Reason: In the presence of the more volatile solute, only the solute will form the vapours and solvent will not.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. Both solute and solvent will form the vapour but vapour phase will be richer in solute.

Teacher's Note:
1. Both components of a liquid mixture contribute to the vapour; the vapour is only richer in the more volatile one.

Question 39. Assertion: If red blood cells were removed from the body and placed in pure water, pressure inside the cells increases.
Reason: The concentration of the salt content in the cells increases. (AIIMS 2006)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (c)
Correct R. Water enters into the cells.

Teacher's Note:
1. Water moves into the cells by osmosis, so the salt inside gets diluted, not concentrated, and the cells swell.