Solutions and Phase Equilibria Questions and Answers

Q: Which of the following aqueous non-electrolyte solution will produce the highest freezing point if 20g of the solute is dissolved in 1000g of water?

The highest freezing point is given by sucrose.Freezing-point depression is a colligative property: ΔTf = Kf × molality, so it depends on the number of solute particles, not their kind. For the same 20 g in 1000 g of water, the solute with the largest molar mass gives the fewest moles, the smallest depression and so the highest freezing point.Sucrose has the largest molar mass of the five, 342 g mol-1, against glucose 180, glycerol 92, ethanol 46 and methanol 32. Methanol, with the smallest molar mass, would give the lowest freezing point.

Q: Isotonic solutions have identical

Isotonic solutions have identical osmotic pressure.Two solutions are isotonic when they have the same osmotic pressure at a given temperature, so no net osmosis takes place between them across a semipermeable membrane. For example, a 0.9% sodium chloride solution is isotonic with blood plasma, which is why it is used in drips.Equal osmotic pressure does not guarantee equal boiling point, freezing point or vapour pressure lowering: these depend on molality rather than molarity and on the solvent, so the definition is stated in terms of osmotic pressure alone.

Q: A binary solution containing two volatile solutes A and B shows positive deviation from Raoult's law. Which of the following statement is true?

The true statement is that the A-B interactions are weaker than the A-A or B-B interactions.In a positive deviation, molecules escape from the mixture more easily than from the pure liquids, so the vapour pressure is higher than Raoult's law predicts. That happens when the attractions between unlike molecules (A-B) are weaker than those between like ones; mixing such liquids absorbs heat and increases the volume. Ethanol and acetone are a standard example.Stronger A-B interactions give a negative deviation, and equal interactions give an ideal solution. The other options describe neither case correctly.

Q: Molal boiling point elevation is the least for the solvent

The molal boiling point elevation is least for water.The molal elevation constant (ebullioscopic constant) Kb is the rise in boiling point for a 1 molal solution. Its values are about 0.52 K kg mol-1 for water, 1.2 for ethanol, 2.53 for benzene, 2.79 for cyclohexane and about 5 for carbon tetrachloride.Water's Kb is the smallest because of its low molar mass and its large enthalpy of vaporisation, so the other solvents all show a larger elevation.

Q: What is the volume of methanol needed for making 2 L of 0.4 M solution? (Density of methanol = 0.64 kgL-1 and molar mass = 32 g mol-1)

The volume needed is 40 mL.Moles of methanol in 2 L of 0.4 M solution, then mass, then volume from the given density (0.64 kg L-1 = 0.64 g mL-1):\[ \begin{aligned} n &= 0.4 \times 2 = 0.8\ \text{mol} \\ m &= 0.8 \times 32 = 25.6\ \text{g} \\ V &= \frac{25.6}{0.64} = 40\ \text{mL} \end{aligned} \]20 mL and 80 mL halve or double the moles (using 1 L or 4 L), and 10 mL or 4 mL come from mixing up kilograms and grams in the density.

Q: What is the mass of ethanoic acid required to prepare 0.5 m solution containing 100 g of be water? (Molar mass of ethanoic acid = 60 g mol-1).

The mass needed is 3 g.Molality is moles of solute per kilogram of solvent. For 100 g (0.1 kg) of water at 0.5 m:\[ \begin{aligned} n &= 0.5 \times 0.1 = 0.05\ \text{mol} \\ m &= 0.05 \times 60 = 3\ \text{g} \end{aligned} \]6 g treats the solution as 0.5 mol per 200 g or doubles the moles, and 0.3 g uses 0.01 kg of water.

Q: 100 cm3 of an aqueous solution of a protein contains 1.5 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 4.5 × 10-3 bar. The molar mass of the protein is[R = 0.083 lit atm mol-1 K-1]

The molar mass is \(8.3 \times 10^{4}\) g mol-1.Use \(\pi = \frac{n}{V}RT\) with \(n = \frac{w}{M}\), so \(M = \frac{wRT}{\pi V}\). The value 0.083 is the gas constant in L bar mol-1 K-1 (the stem labels it lit atm), which suits the pressure given in bar; V = 0.1 L.\[ \begin{aligned} M &= \frac{1.5 \times 0.083 \times 300}{4.5 \times 10^{-3} \times 0.1} \\ &= \frac{37.35}{4.5 \times 10^{-4}} \\ &= 8.3 \times 10^{4}\ \text{g mol}^{-1} \end{aligned} \]\(8.3 \times 10^{5}\) and \(8.3 \times 10^{3}\) come from taking the volume as 0.01 L or 1 L. The two options showing \(4.15 \times 10^{4}\), half the answer, are identical in the paper.

Q: The salt which has van't Hoff factor (i) as 1.82 in 0.001 m aqueous solution is

The salt is MgSO4.MgSO4 gives two ions, so its ideal van't Hoff factor is 2. Interionic attraction between the doubly charged Mg2+ and SO42- ions is strong, so even at 0.001 m the observed factor is lower, 1.82 in the textbook table.KCl and NaCl also give two ions, but their singly charged ions attract less, so their factor at 0.001 m is about 1.94. K2SO4 gives three ions, with a factor near 2.84. HCl, a strong acid, is close to 2 at this dilution.

Q: The mass of sodium acetate required to prepare 250 mL of 0.4 M aqueous solution is (Molar mass of sodium acetate = 82 g mol-1)

The mass required is 8.2 g.Moles needed = molarity times volume in litres; mass = moles times molar mass:\[ \begin{aligned} n &= 0.4 \times 0.250 = 0.1\ \text{mol} \\ m &= 0.1 \times 82 = 8.2\ \text{g} \end{aligned} \]4.1 g and 16.4 g are half and double the right amount, as from using 125 mL or 500 mL. 0.82 g is ten times too small.

Q: 3 g of benzoic acid dissolved in 25 g of benzene shows a depression in freezing point equal to 2.5 K. What is the experimental molar mass of benzoic acid in benzene? (Molal depression constant for benzene is 5 K kg mol-1).

The experimental molar mass is 240 g mol-1.From \(\Delta T_{f} = K_{f}m\) with \(m = \frac{w_{2} \times 1000}{M_{2} \times w_{1}}\):\[ \begin{aligned} M_{2} &= \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} \\ &= \frac{5 \times 3 \times 1000}{2.5 \times 25} = 240\ \text{g mol}^{-1} \end{aligned} \]This is about twice the true molar mass of benzoic acid, 122 g mol-1, because in benzene benzoic acid forms hydrogen-bonded dimers, halving the number of particles. 122 is the formula value, not the experimental one, and 244 is exactly twice it.

21.

Which of the following aqueous non-electrolyte solution will produce the highest freezing point if 20g of the solute is dissolved in 1000g of water?

A.Sucrose
B.Glycerol
C.Ethanol
D.Glucose
E.Methanol

Exam Year(s)

KEAM - 2026,

Answer: A

The highest freezing point is given by sucrose.

Freezing-point depression is a colligative property: ΔT_f = K_f × molality, so it depends on the number of solute particles, not their kind. For the same 20 g in 1000 g of water, the solute with the largest molar mass gives the fewest moles, the smallest depression and so the highest freezing point.

Sucrose has the largest molar mass of the five, 342 g mol^-1, against glucose 180, glycerol 92, ethanol 46 and methanol 32. Methanol, with the smallest molar mass, would give the lowest freezing point.

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22.

Isotonic solutions have identical

A.boiling point
B.freezing point
C.vapour pressure
D.osmotic pressure
E.lowering of vapour pressure

Exam Year(s)

KEAM - 2026,

Answer: D

Isotonic solutions have identical osmotic pressure.

Two solutions are isotonic when they have the same osmotic pressure at a given temperature, so no net osmosis takes place between them across a semipermeable membrane. For example, a 0.9% sodium chloride solution is isotonic with blood plasma, which is why it is used in drips.

Equal osmotic pressure does not guarantee equal boiling point, freezing point or vapour pressure lowering: these depend on molality rather than molarity and on the solvent, so the definition is stated in terms of osmotic pressure alone.

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23.

A binary solution containing two volatile solutes A and B shows positive deviation from Raoult's law. Which of the following statement is true?

A.The molecular interactions between A-B are greater than those of A-A or B-B.
B.The molecular interactions between A-B are weaker than those of A-A or B-B.
C.There are equal molecular interactions between A-A or B-B.
D.There are no molecular interactions between A-B compared to those of A-A and B-B.
E.The molecular interactions between one of the components is stronger than those of the other component.

Exam Year(s)

KEAM - 2026,

Answer: B

The true statement is that the A-B interactions are weaker than the A-A or B-B interactions.

In a positive deviation, molecules escape from the mixture more easily than from the pure liquids, so the vapour pressure is higher than Raoult's law predicts. That happens when the attractions between unlike molecules (A-B) are weaker than those between like ones; mixing such liquids absorbs heat and increases the volume. Ethanol and acetone are a standard example.

Stronger A-B interactions give a negative deviation, and equal interactions give an ideal solution. The other options describe neither case correctly.

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24.

Molal boiling point elevation is the least for the solvent

A.water
B.ethanol
C.benzene
D.carbon tetrachloride
E.cyclohexane

Exam Year(s)

KEAM - 2026,

Answer: A

The molal boiling point elevation is least for water.

The molal elevation constant (ebullioscopic constant) K_b is the rise in boiling point for a 1 molal solution. Its values are about 0.52 K kg mol^-1 for water, 1.2 for ethanol, 2.53 for benzene, 2.79 for cyclohexane and about 5 for carbon tetrachloride.

Water's K_b is the smallest because of its low molar mass and its large enthalpy of vaporisation, so the other solvents all show a larger elevation.

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25.

What is the volume of methanol needed for making 2 L of 0.4 M solution? (Density of methanol = 0.64 kgL^-1 and molar mass = 32 g mol^-1)

A.20 mL
B.4 mL
C.40 mL
D.10 mL
E.80 mL

Exam Year(s)

KEAM - 2026,

Answer: C

The volume needed is 40 mL.

Moles of methanol in 2 L of 0.4 M solution, then mass, then volume from the given density (0.64 kg L^-1 = 0.64 g mL^-1):

\[ \begin{aligned} n &= 0.4 \times 2 = 0.8\ \text{mol} \\ m &= 0.8 \times 32 = 25.6\ \text{g} \\ V &= \frac{25.6}{0.64} = 40\ \text{mL} \end{aligned} \]

20 mL and 80 mL halve or double the moles (using 1 L or 4 L), and 10 mL or 4 mL come from mixing up kilograms and grams in the density.

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26.

What is the mass of ethanoic acid required to prepare 0.5 m solution containing 100 g of be water? (Molar mass of ethanoic acid = 60 g mol^-1).

A.3 g
B.6 g
C.0.3 g
D.7.5 g
E.2 g

Exam Year(s)

KEAM - 2026,

Answer: A

The mass needed is 3 g.

Molality is moles of solute per kilogram of solvent. For 100 g (0.1 kg) of water at 0.5 m:

\[ \begin{aligned} n &= 0.5 \times 0.1 = 0.05\ \text{mol} \\ m &= 0.05 \times 60 = 3\ \text{g} \end{aligned} \]

6 g treats the solution as 0.5 mol per 200 g or doubles the moles, and 0.3 g uses 0.01 kg of water.

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27.

100 cm³ of an aqueous solution of a protein contains 1.5 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 4.5 × 10^-3 bar. The molar mass of the protein is
[R = 0.083 lit atm mol^-1 K^-1]

A.\(8.3 \times 10^{5}\ \mathrm{g\ mol^{-1}}\)
B.\(4.15 \times 10^{4}\ \mathrm{g\ mol^{-1}}\)
C.\(8.3 \times 10^{3}\ \mathrm{g\ mol^{-1}}\)
D.\(8.3 \times 10^{4}\ \mathrm{g\ mol^{-1}}\)
E.\(4.15 \times 10^{4}\ \mathrm{g\ mol^{-1}}\)

Exam Year(s)

KEAM - 2026,

Answer: D

The molar mass is \(8.3 \times 10^{4}\) g mol^-1.

Use \(\pi = \frac{n}{V}RT\) with \(n = \frac{w}{M}\), so \(M = \frac{wRT}{\pi V}\). The value 0.083 is the gas constant in L bar mol^-1 K^-1 (the stem labels it lit atm), which suits the pressure given in bar; V = 0.1 L.

\[ \begin{aligned} M &= \frac{1.5 \times 0.083 \times 300}{4.5 \times 10^{-3} \times 0.1} \\ &= \frac{37.35}{4.5 \times 10^{-4}} \\ &= 8.3 \times 10^{4}\ \text{g mol}^{-1} \end{aligned} \]

\(8.3 \times 10^{5}\) and \(8.3 \times 10^{3}\) come from taking the volume as 0.01 L or 1 L. The two options showing \(4.15 \times 10^{4}\), half the answer, are identical in the paper.

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28.

The salt which has van't Hoff factor (i) as 1.82 in 0.001 m aqueous solution is

A.\(\mathrm{K_{2}SO_{4}}\)
B.\(\mathrm{MgSO_{4}}\)
C.KCl
D.NaCl
E.HCl

Exam Year(s)

KEAM - 2026,

Answer: B

The salt is MgSO₄.

MgSO₄ gives two ions, so its ideal van't Hoff factor is 2. Interionic attraction between the doubly charged Mg²⁺ and SO₄^2- ions is strong, so even at 0.001 m the observed factor is lower, 1.82 in the textbook table.

KCl and NaCl also give two ions, but their singly charged ions attract less, so their factor at 0.001 m is about 1.94. K₂SO₄ gives three ions, with a factor near 2.84. HCl, a strong acid, is close to 2 at this dilution.

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29.

The mass of sodium acetate required to prepare 250 mL of 0.4 M aqueous solution is (Molar mass of sodium acetate = 82 g mol^-1)

A.4.1 g
B.8.2 g
C.16.4 g
D.2.05 g
E.0.82 g

Exam Year(s)

KEAM - 2026,

Answer: B

The mass required is 8.2 g.

Moles needed = molarity times volume in litres; mass = moles times molar mass:

\[ \begin{aligned} n &= 0.4 \times 0.250 = 0.1\ \text{mol} \\ m &= 0.1 \times 82 = 8.2\ \text{g} \end{aligned} \]

4.1 g and 16.4 g are half and double the right amount, as from using 125 mL or 500 mL. 0.82 g is ten times too small.

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30.

3 g of benzoic acid dissolved in 25 g of benzene shows a depression in freezing point equal to 2.5 K. What is the experimental molar mass of benzoic acid in benzene? (Molal depression constant for benzene is 5 K kg mol-¹).

A.122 g \(\mathrm{mol^{-1}}\)
B.244 g \(\mathrm{mol^{-1}}\)
C.240 g \(\mathrm{mol^{-1}}\)
D.242 g \(\mathrm{mol^{-1}}\)
E.250 g \(\mathrm{mol^{-1}}\)

Exam Year(s)

KEAM - 2026,

Answer: C

The experimental molar mass is 240 g mol^-1.

From \(\Delta T_{f} = K_{f}m\) with \(m = \frac{w_{2} \times 1000}{M_{2} \times w_{1}}\):

\[ \begin{aligned} M_{2} &= \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} \\ &= \frac{5 \times 3 \times 1000}{2.5 \times 25} = 240\ \text{g mol}^{-1} \end{aligned} \]

This is about twice the true molar mass of benzoic acid, 122 g mol^-1, because in benzene benzoic acid forms hydrogen-bonded dimers, halving the number of particles. 122 is the formula value, not the experimental one, and 244 is exactly twice it.

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