Solutions - MCQs with Explanations

1. Which of the following does not show positive deviation from Raoult's law?

• (a) Benzene-chloroform
• (b) Benzene-acetone
• (c) Benzene-ethanol
• (d) Benzene-carbon tetrachloride

Correct Answer: (a)

Explanation: A mixture of benzene and chloroform shows negative deviation from Raoult's law.

2. Which of the following aqueous solutions have higher freezing point?

• (a) 0.1 m \(Al_{2}(SO_{4})_{3}\)
• (b) 0.1 m \(BaCl_{2}\)
• (c) 0.1 m \(AlCl_{3}\)
• (d) 0.1 m \(NH_{4}Cl\)

Correct Answer: (d)

Explanation: The freezing point is higher when the depression in freezing point is smaller. This occurs with a lesser number of particles in the solution. Among the options, \(NH_{4}Cl\) produce the fewest particles.

3. A glucose solution is to be injected into the blood stream, it must have the same ________ as the blood stream.

• (a) molarity
• (b) vapour pressure
• (c) osmotic pressure
• (d) viscosity

Correct Answer: (c)

Explanation: Solutions intended for injection must be isotonic with blood, meaning they must have the same osmotic pressure.

4. If an equimolar solution of \(CaCl_{2}\) and \(AlCl_{3}\) in water have boiling point of \(T_{1}\) and \(T_{2}\) respectively then:

• (a) \(T_{1} > T_{2}\)
• (b) \(T_{2} > T_{1}\)
• (c) \(T_{1} = T_{2}\)
• (d) \(T_{1} \neq T_{2}\)

Correct Answer: (b)

Explanation: \(AlCl_{3}\) dissociates into more particles than \(CaCl_{2}\). Consequently, the elevation in boiling point (\(\Delta T_{b}\)) is higher for \(AlCl_{3}\), resulting in a higher boiling point \(T_{2}\).

5. Vapour pressure of \(CCl_{4}\) at \(25^{\circ}C\) is 143 mm Hg. 0.5 g of a non-volatile solute (mol. weight 65) is dissolved in 100 mL of \(CCl_{4}\). Find the vapour pressure of the solution (Density of \(CCl_{4} = 1.58 g/cm^{3}\)).

• (a) 141.93 mm
• (b) 94.39 mm
• (c) 199.34 mm
• (d) 143.99 mm

Correct Answer: (a)

Explanation: Using the formula for relative lowering of vapour pressure, the pressure of the solution is calculated as approximately 141.93 mm.

6. Which of the following is incorrect?

• (a) Molecular weight of NaCl found by osmotic pressure measurements is half of the theoretical value.
• (b) Molecular weight of \(CH_{3}COOH\) in benzene found by cryoscopic methods is double of the theoretical value.
• (c) Osmotic pressure of 0.1 M glucose solution is half of that of 0.1 M NaCl solution.
• (d) Molecular weight of HCl found by any colligative property will be same in the aqueous solution and benzene solution.

Correct Answer: (d)

Explanation: This is incorrect because HCl dissociates into ions in aqueous solution (affecting observed molecular weight) but does not dissociate in benzene.

7. The molar freezing point constant for water is \(1.86^{\circ}C/m\). If 342 g of cane sugar (\(C_{12}H_{22}O_{11}\)) is dissolved in 1000 g of water, the solution will freeze at:

• (a) \(-1.86^{\circ}C\)
• (b) \(1.86^{\circ}C\)
• (c) \(-3.92^{\circ}C\)
• (d) \(2.42^{\circ}C\)

Correct Answer: (a)

Explanation: 342 g of sugar in 1000 g water is a 1 molal solution. The depression (\(\Delta T_{f}\)) is \(1.86^{\circ}C\), so the freezing point is \(-1.86^{\circ}C\).

8. For an ideal solution with \(p_{A}^{\circ} > p_{B}^{\circ}\), which of the following is true?

• (a) \((x_{A})_{liquid} = \((x_{A})_{vapour}\)
• (b) \((x_{A})_{liquid} > \((x_{A})_{vapour}\)
• (c) \((x_{A})_{liquid} < \((x_{A})_{vapour}\)
• (d) There is no relationship between \((x_{A})_{liquid}\) and \((x_{A})_{vapour}\).

Correct Answer: (c)

Explanation: Since component A is more volatile, the vapour phase will be richer in component A than the liquid phase.

9. The vapour pressures of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is:

• (a) 0.50
• (b) 0.16
• (c) 0.27
• (d) 0.73

Correct Answer: (c)

Explanation: For an equimolar solution, partial pressures are 80 torr (benzene) and 30 torr (toluene). Mole fraction of toluene in vapour = \(30 / (80 + 30) = 0.27\).

10. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (molar mass = 342). The molality of the solution is:

• (a) 0.0099
• (b) 0.56
• (c) 0.28
• (d) 0.34

Correct Answer: (b)

Explanation: Moles of sugar = 0.1. Mass of water = \(214.2 - 34.2 = 180 g = 0.18 kg\). Molality = \(0.1 / 0.18 = 0.56 m\).

11. 45 g of ethylene glycol (\(C_{2}H_{6}O_{2}\)) is mixed with 600 g of water. The freezing point of the solution is (\(K_{f}\) for water is \(1.86 K kg mol^{-1}\)):

• (a) 273.95 K
• (b) 270.95 K
• (c) 370.95 K
• (d) 373.95 K

Correct Answer: (b)

Explanation: The molality is 1.2 mol/kg, leading to a depression (\(\Delta T_{f}\)) of 2.2 K. The freezing point is 270.95 K.

12. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol\(^{-1}\). The molar mass of the solute is:

• (a) 256 kg/mol
• (b) 256 g mol
• (c) 256 g/mol
• (d) 256 mg/mol

Correct Answer: (c)

Explanation: Using the depression formula, the molar mass is calculated to be 256 g/mol.

13. The molal freezing point depression constant for benzene (\(C_{6}H_{6}\)) is \(4.90 K kg mol^{-1}\). Selenium exists as a polymer of the type \(Se_{x}\). When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is \(0.112^{\circ}C\) lower than that of pure benzene. The molecular formula of selenium is (Atomic mass of \(Se = 78.8 g mol^{-1}\)):

• (a) \(Se_{4}\)
• (b) \(Se_{2}\)
• (c) \(Se_{6}\)
• (d) \(Se_{8}\)

Correct Answer: (d)

Explanation: The calculated molecular weight is 631.08. Dividing this by the atomic mass (78.8) gives \(x = 8\), hence \(Se_{8}\).

14. \(FeCl_{3}\) on reaction with \(K_{4}[Fe(CN)_{6}]\) in aqueous solution gives blue colour. These two solutions are separated by a semipermeable membrane XY. Due to osmosis there is:

• (a) blue colour formation in side A
• (b) blue colour formation in side B
• (c) blue colour on both the sides
• (d) no blue colour formation.

Correct Answer: (d)

Explanation: In osmosis, only the solvent (water) molecules pass through the membrane. The solute ions cannot pass, so no reaction (and no blue colour) occurs.

15. Which one of the following statements is false?

• (a) Raoult's law states that the vapour pressure of a component over a binary solution of volatile liquids is directly proportional to its mole fraction.
• (b) Two sucrose solutions of the same molality prepared in different solvents will have the same depression of freezing point.
• (c) The correct order of osmotic pressures of 0.01 M solution of each compound is \(BaCl_{2} > KCl > CH_{3}COOH > glucose\).
• (d) In the equation osmotic pressure \(\pi = MRT\), M is the molarity of the solution.

Correct Answer: (b)

Explanation: This is false because the depression constant \(K_{f}\) is specific to each individual solvent.

16. Solubility of a gas in a liquid increases with:

• (a) increase of pressure and increase of temperature
• (b) decrease of pressure and increase of temperature
• (c) increase of pressure and decrease of temperature
• (d) decrease of pressure and decrease of temperature.

Correct Answer: (c)

Explanation: Solubility increases with increased pressure (Henry's law) and decreased temperature.

17. The relationship between osmotic pressure at 273 K when 10 g glucose (\(p_{1}\)), 10 g urea (\(p_{2}\)) and 10 g sucrose (\(p_{3}\)) are dissolved in 250 mL of water, is:

• (a) \(p_{1} > p_{2} > p_{3}\)
• (b) \(p_{3} > p_{1} > p_{2}\)
• (c) \(p_{2} > p_{1} > p_{3}\)
• (d) \(p_{2} > p_{3} > p_{1}\)

Correct Answer: (c)

Explanation: Higher molecular weight results in lower moles and lower osmotic pressure for the same mass. Order of MW: sucrose > glucose > urea. Therefore, order of pressure: urea > glucose > sucrose.

18. Boiling point of water at 750 mm Hg is \(96.63^{\circ}C\). How much sucrose is to be added to 500 g of water such that it boils at \(100^{\circ}C\)? Molal elevation constant for water is \(0.52 K kg mol^{-1}\).

• (a) 11.08 kg
• (b) 1108.2 g
• (c) 1108.2 kg
• (d) 11.08 g

Correct Answer: (b)

Explanation: Using the boiling point elevation formula, the mass of sucrose required is calculated as 1108.2 g.

19. Liquids A and B form an ideal solution and the former has stronger intermolecular forces. If \(x_{A}\) and \(x'_{A}\) are the mole fractions of A in the solution and vapour in equilibrium, then:

• (a) \(x'_{A}/x_{A} = 1\)
• (b) \(x'_{A}/x_{A} > 1\)
• (c) \(x'_{A}/x_{A} < 1\)
• (d) \(x'_{A} + x_{A} = 1\)

Correct Answer: (c)

Explanation: Since A is less volatile (stronger forces), it will be less concentrated in the vapour phase compared to the liquid phase, making the ratio less than 1.

20. The percentage dissociation of a 0.011 m aqueous solution of \(K_{3}[Fe(CN)_{6}]\) which freezes at \(-0.063^{\circ}C\) is (\(K_{f}\) for water = 1.86):

• (a) 75%
• (b) 67%
• (c) 33%
• (d) 50%

Correct Answer: (b)

Explanation: The observed van't Hoff factor is 3. With \(n = 4\), the degree of dissociation is 0.67 or 67%.

21. A motor vehicle radiator was filled with 8 litre of water in which 2 litre of methyl alcohol was added. What is the lowest temperature at which the vehicle can be parked out without a danger of freezing of water? (\(K_{f}\) for water = \(1.86 K kg mol^{-1}\) and density of methanol = \(0.8 g/mL\)).

• (a) 11.625
• (b) -11.625
• (c) 14.531
• (d) -14.531

Correct Answer: (b)

Explanation: The calculated depression (\(\Delta T_{f}\)) is 11.625, meaning the freezing point is \(-11.625^{\circ}C\).

22. What is the freezing point of a solution containing 8.1 g HBr in 100 g water, assuming that the acid to be 90% ionized? (\(K_{f}(H_{2}O) = 1.86 K kg mol^{-1}\)).

• (a) \(0.85^{\circ}C\)
• (b) \(-3.8^{\circ}C\)
• (c) \(-0^{\circ}C\)
• (d) \(-3.5^{\circ}C\)

Correct Answer: (d)

Explanation: The van't Hoff factor for 90% ionization is 1.90. The depression is calculated as 3.5 K, so the freezing point is \(-3.5^{\circ}C\).

23. Vapour pressure of \(C_{6}H_{6}\) and \(C_{7}H_{8}\) mixture at \(50^{\circ}C\) are given by: \(P_{total} = 179 X_{B} + 92\) (\(X_{B} = \) mole fraction of \(C_{6}H_{6}\)). Then the vapour pressure of pure benzene in mm of Hg is:

• (a) 271
• (b) 92
• (c) 380
• (d) 120

Correct Answer: (a)

Explanation: Comparison with Raoult's law formula shows \(p_{T}^{\circ} = 92\) and \(p_{B}^{\circ} - p_{T}^{\circ} = 179\). Thus, \(p_{B}^{\circ} = 179 + 92 = 271 mm Hg\).

24. If 0.5 m solution of \(Ca(NO_{3})_{2}\) and 0.75 m solution of KOH is taken, then the depression in freezing point is:

• (a) greater in \(Ca(NO_{3})_{2}\) because number of ions are greater
• (b) greater in KOH because concentration is high
• (c) equal in both and freezing point is less than \(0^{\circ}C\) because ionic concentration is same
• (d) equal to \(0^{\circ}C\) in both because ionic concentration is negligible.

Correct Answer: (c)

Explanation: Both solutions have an identical ionic concentration of 1.50 particles (0.5 x 3 for calcium nitrate; 0.75 x 2 for KOH), leading to equal depression.

25. 4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molarity of the resultant solution is:

• (a) 0.004
• (b) 0.008
• (c) 0.012
• (d) 0.016

Correct Answer: (d)

Explanation: Final volume = 5 L. Using \(M_{1}V_{1} = M_{2}V_{2}\), the final molarity is \((0.02 \times 4) / 5 = 0.016 M\).

26. The van't Hoff factor for dilute solution of glucose is:

• (a) zero
• (b) 1
• (c) 1.5
• (d) 2.0

Correct Answer: (b)

Explanation: Glucose is a non-electrolyte, so its van't Hoff factor is 1.

27. Two elements A and B form compounds having molecular formulae \(AB_{2}\) and \(AB_{4}\), when dissolved in 20 g of \(C_{6}H_{6}\) 1.0 g of \(AB_{2}\) lowers the freezing point by 2.3 K whereas 1.0 g of \(AB_{4}\) lowers it by 1.3 K. The molal depression constant for benzene is 5.1 K kg mol\(^{-1}\). The atomic masses of A and B are, respectively:

• (a) 26, 42.64
• (b) 31.72, 47.02
• (c) 13.11, 24.25
• (d) 19.17, 35.01

Correct Answer: (a)

Explanation: Solving the depression equations for both compounds yields atomic masses of approximately 26 (A) and 42.64 (B).

28. At the same temperature which of the following solutions will be isotonic?

• (a) 3.42 g of sucrose per litre of water and 0.18 g of glucose per litre of water.
• (b) 3.42 g of sucrose per litre of water and 0.18 g of glucose in 0.1 litre of water.
• (c) 3.42 g of sucrose per litre of water and 0.585 g of sodium chloride per litre of water.
• (d) 3.42 g of sucrose per litre of water and 1.17 g of sodium chloride per litre of water.

Correct Answer: (b)

Explanation: Both solutions in option (b) have a molar concentration of 0.01 M, making them isotonic.

29. 0.85%, aqueous solution of \(NaNO_{3}\) is apparently 90% dissociated at \(27^{\circ}C\). The osmotic pressure will be (R = 0.082 atm K\(^{-1}\) mol\(^{-1}\)):

• (a) 2.210 atm
• (b) 4.674 atm
• (c) 3.049 atm
• (d) 5.012 atm

Correct Answer: (b)

Explanation: With 90% dissociation, the van't Hoff factor is 1.9. The osmotic pressure is calculated as 4.674 atm.

30. The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows:

• (a) negative deviations from Raoult's law
• (b) positive deviations from Raoult's law
• (c) no deviation from Raoult's law
• (d) deviations which can not be predicted from the given information.

Correct Answer: (b)

Explanation: A minimum boiling azeotrope occurs when the mixture shows positive deviation from Raoult's law.

31. At a particular temperature, the vapour pressure of two liquids R and Q are respectively 120 and 180 mm of mercury. If 2 moles of R and 3 moles of Q are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury):

• (a) 156
• (b) 145
• (c) 150
• (d) 108

Correct Answer: (a)

Explanation: Total pressure \(P = (0.4 \times 120) + (0.6 \times 180) = 156 mm Hg\).

32. The Henry's law constant for the solubility of \(N_{2}\) gas in water at 298 K is \(1.0 \times 10^{5}\) atm. The mole fraction of \(N_{2}\) in air is 0.8. The number of moles of \(N_{2}\) from air dissolved in 10 moles of water at 298 K and 5 atm pressure is:

• (a) \(4.0 \times 10^{-4}\)
• (b) \(4.0 \times 10^{-5}\)
• (c) \(5.0 \times 10^{-4}\)
• (d) \(4.0 \times 10^{-6}\)

Correct Answer: (a)

Explanation: Using Henry's law, the calculated number of moles dissolved in 10 moles of water is \(4.0 \times 10^{-4}\).

33. 12 g of urea is dissolved in 1 L of water and 68.4 g of sucrose is dissolved in 1 L of water. The lowering of vapour pressure of urea is:

• (a) equal to sucrose
• (b) greater than sucrose
• (c) less than sucrose
• (d) double that of sucrose

Correct Answer: (a)

Explanation: Both solutions contain 0.2 moles of solute in the same volume of solvent, resulting in the same mole fraction and identical lowering of vapour pressure.

34. Calculate the molality of a 1 L solution of 93% \(H_{2}SO_{4}\) (mass/volume). The density of the solution is \(1.84 g/mL\).

• (a) 9.48 m
• (b) 10.30 m
• (c) 11.33 m
• (d) 12.06 m

Correct Answer: (b)

Explanation: The molarity is 9.48 M. Converting this to molality using the density yields 10.30 m.

35. Pure water boils at \(99.725^{\circ}C\) at Shimla. If \(K_{b}\) for water is \(0.51^{\circ}C molal^{-1}\), the boiling point of 0.69 molal urea solution will be:

• (a) \(100.35^{\circ}C\)
• (b) \(100.08^{\circ}C\)
• (c) \(99.37^{\circ}C\)
• (d) none of these

Correct Answer: (b)

Explanation: The calculated elevation is 0.3519. The new boiling point is \(99.725 + 0.3519 = 100.08^{\circ}C\).

36. The van't Hoff factor for 0.1 M \(Ba(NO_{3})_{2}\) solution is 2.74. The degree of dissociation is:

• (a) 73.4%
• (b) 87%
• (c) 100%
• (d) 74%

Correct Answer: (b)

Explanation: Using the formula \(\alpha = (i - 1) / (n - 1)\) with \(n = 3\), the degree of dissociation is 0.87 or 87%.

37. Calculate the mole fraction of ethylene glycol (\(C_{2}H_{6}O_{2}\)) in a solution containing 20% of \(C_{2}H_{6}O_{2}\) by mass.

• (a) 0.932
• (b) 0.068
• (c) 0.950
• (d) 0.779

Correct Answer: (b)

Explanation: The moles of glycol and water lead to a mole fraction of 0.068 for ethylene glycol.

38. Vapour pressure of benzene at \(30^{\circ}C\) is 121.8 mm. When 15 g of a non-volatile solute is dissolved in 250 g of benzene, its vapour pressure is decreased to 120.2 mm. The molecular weight of the solute is:

• (a) 35.67 g
• (b) 357.8 g
• (c) 432.8 g
• (d) 502.7 g

Correct Answer: (b)

Explanation: The molecular weight is calculated as 357.8 g using the relative lowering formula.

39. The degree of dissociation (\(\alpha\)) of a weak electrolyte \(A_{x}B_{y}\) is related to van't Hoff factor (i) by the expression:

• (a) \(\alpha = (i - 1) / (x + y - 1)\)
• (b) \(\alpha = (i - 1) / (x + y + 1)\)
• (c) \(\alpha = (x + y - 1) / (i - 1)\)
• (d) \(\alpha = (x + y + 1) / (i - 1)\)

Correct Answer: (a)

Explanation: This matches the standard derivation where the total number of ions produced is \(n = x + y\).

40. Solubility curve of \(Na_{2}SO_{4}\cdot 10H_{2}O\) in water with temperature is given. From the figure, we can say that:

• (a) solution process is exothermic
• (b) solution process is exothermic till \(34^{\circ}C\) and endothermic after \(34^{\circ}C\)
• (c) solution process is endothermic till \(34^{\circ}C\) and exothermic thereafter
• (d) solution process is endothermic.

Correct Answer: (c)

Explanation: The dissolution of the decahydrate is endothermic until \(34^{\circ}C\); thereafter, the anhydrous form dissolves exothermically.

41. The order of increasing freezing point of \(C_{2}H_{5}OH\), \(Ba_{3}(PO_{4})_{2}\), \(Na_{2}SO_{4}\), \(KCl\) and \(Li_{3}PO_{4}\) is:

• (a) \(Ba_{3}(PO_{4})_{2} < Na_{2}SO_{4} < Li_{3}PO_{4} < C_{2}H_{5}OH < KCl\)
• (b) \(Ba_{3}(PO_{4})_{2} < C_{2}H_{5}OH < Li_{3}PO_{4} < Na_{2}SO_{4} < KCl\)
• (c) \(C_{2}H_{5}OH < KCl < Na_{2}SO_{4} < Ba_{3}(PO_{4})_{2} < Li_{3}PO_{4}\)
• (d) \(Ba_{3}(PO_{4})_{2} < Li_{3}PO_{4} < Na_{2}SO_{4} < KCl < C_{2}H_{5}OH\)

Correct Answer: (d)

Explanation: Freezing point increases as the van't Hoff factor (i) decreases. The order of \(i\) values is 5, 4, 3, 2, 1, matching the sequence in (d).

42. Phenol associates in benzene to a certain extent to form dimer. A solution containing \(2.0 \times 10^{-2}\) kg of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is (\(K_{f}\) for benzene = \(5.12 K kg mol^{-1}\)):

• (a) 73.4%
• (b) 50.1%
• (c) 100%
• (d) 25.1%

Correct Answer: (a)

Explanation: The observed van't Hoff factor is 0.633. The degree of association is calculated as 0.734 or 73.4%.

43. When 0.6 g of an organic acid dissolved in 200 mL of water is extracted with 50 mL of ether, it is found on analysis, that 0.4 g of the organic acid goes into the ether layer. Which of the following represents the correct ratio of the solubility of the acid in ether and water?

• (a) \(S_{ether}/S_{water} = 0.125\)
• (b) \(S_{ether}/S_{water} = 0.5\)
• (c) \(S_{ether}/S_{water} = 2\)
• (d) \(S_{ether}/S_{water} = 8\)

Correct Answer: (d)

Explanation: Solubility in water = 0.2g/200mL. Solubility in ether = 0.4g/50mL (or 1.6g/200mL). Ratio = 1.6 / 0.2 = 8.

44. A 3% aqueous solution of urea is isotonic to 5% aqueous solution of another solute. Considering van't Hoff factor of the other solute to be 1.0, determine its molecular mass.

• (a) 50
• (b) 60
• (c) 90
• (d) 100

Correct Answer: (d)

Explanation: For isotonicity, \(n_{1} = n_{2}\). Thus \(3 / 60 = 5 / Mol.mass\), which gives a molecular mass of 100.

45. Match the terms given in Column I with the type of solutions given in Column II:

A. Soda water; B. Sugar solution; C. German silver; D. Air; E. Hydrogen gas in palladium.

• (a) A-5, B-6, C-4, D-2, E-1
• (b) A-5, B-3, C-4, D-2, E-1
• (c) A-1, B-2, C-3, D-4, E-5
• (d) A-1, B-2, C-3, D-4, E-6

Correct Answer: (b)

Explanation: Soda water is gas in liquid; sugar solution is solid in liquid; German silver is solid in solid; Air is gas in gas; Hydrogen in palladium is gas in solid.