Colligative Properties – Colligative properties of dilute solutions: NEET MCQs with Answers

15 free questions on Colligative Properties – Colligative properties of dilute solutions for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Colligative properties of dilute solutions - a relative lowering of vapour pressure, depression of freezing point, the elevation of boiling point and osmotic pressure; Determination of molecular mass using colligative properties;

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Question 1

Previous-year question
Which observation(s) reflect(s) colligative properties? (i) A 0.5 m NaBr solution has a higher vapour pressure than a 0.5 m BaCl₂ solution at the same temperature. (ii) Pure water freezes at a higher temperature than pure methanol. (iii) A 0.1 m NaOH solution freezes at a lower temperature than pure water. [NEET 2017]
  1. (i), (ii) and (iii)
  2. (i) and (ii)
  3. (ii) and (iii)
  4. (i) and (iii)
Show answer
Answer: (D) (i) and (iii)
Explanation: Observations (i) and (iii) both compare the effect of the number of solute particles, which is exactly what colligative properties measure. Statement (ii) simply compares two pure liquids and has nothing to do with a solute.

Question 2

Previous-year question
If the molality of a dilute solution is doubled, the value of the molal depression constant (K_f) will be [NEET 2017]
  1. halved
  2. tripled
  3. unchanged
  4. doubled
Show answer
Answer: (C) unchanged
Explanation: K_f is a characteristic constant of the solvent, defined as the depression produced by a one-molal solution. It does not depend on the concentration or on the nature of the solute.

Question 3

Previous-year question
At 100 °C the vapour pressure of a solution of 6.5 g of a solute in 100 g of water is 732 mm. If K_b = 0.52, the boiling point of this solution will be [NEET 2016]
  1. 101 °C
  2. 100 °C
  3. 102 °C
  4. 103 °C
Show answer
Answer: (A) 101 °C
Explanation: Relative lowering = (760 − 732)/760 = 0.0368, and moles of water = 5.56, so n(solute) = 0.205 and m = 2.05 mol kg⁻¹. Hence ΔT_b = 0.52 × 2.05 ≈ 1 °C and the boiling point is 101 °C.

Question 4

Previous-year question
The boiling point of a 0.2 mol kg⁻¹ solution of X in water is greater than that of an equimolal solution of Y in water. Which one of the following statements is true in this case? [AIPMT 2015]
  1. The molecular mass of X is greater than the molecular mass of Y.
  2. The molecular mass of X is less than the molecular mass of Y.
  3. Y is undergoing dissociation in water while X undergoes no change.
  4. X is undergoing dissociation in water.
Show answer
Answer: (D) X is undergoing dissociation in water.
Explanation: Both solutions have the same molality, so the larger elevation for X means it produces more particles in solution. X must therefore be dissociating; molar mass is irrelevant because molality is already fixed.

Question 5

The osmotic pressure method is particularly useful for finding the molar masses of proteins and polymers. Which of the following is NOT a reason for this?
  1. A larger solute molecule exerts a greater osmotic pressure per mole, so macromolecules give larger readings
  2. Its magnitude is large even for very dilute solutions, unlike that of the other colligative properties
  3. Polymers have poor solubility, and the dilute solutions they give can still be studied by this method
  4. The measurement is made at about room temperature, so biomolecules that are unstable at higher temperatures are not damaged
Show answer
Answer: (A) A larger solute molecule exerts a greater osmotic pressure per mole, so macromolecules give larger readings
Explanation: Osmotic pressure is a colligative property given by π = CRT, so it depends on the number of solute particles per unit volume and not on their size. Macromolecules therefore give no extra pressure per mole. The method is favoured because it works around room temperature, which suits heat-sensitive biomolecules, and because π is large enough to measure even in the very dilute solutions that polymers form.

Question 6

In which of the following situations is there a net movement of water OUT of a living cell by osmosis?
  1. A limp carrot is placed in fresh water
  2. Wilted flowers are placed in fresh water
  3. Human blood cells are placed in 0.5% (mass/volume) NaCl solution
  4. A bacterium lands on salted meat or on candied fruit
Show answer
Answer: (D) A bacterium lands on salted meat or on candied fruit
Explanation: Water flows through a semipermeable membrane from the side of lower solute concentration to the side of higher solute concentration. The salt or sugar around a bacterium makes its surroundings hypertonic, so it loses water, shrivels and dies; this is why salting and sugaring preserve food. Fresh water is hypotonic to carrot and flower cells, so water enters and they become firm again. Blood cells have an internal fluid equivalent to 0.9% (mass/volume) NaCl, so they take in water and swell in 0.5% NaCl.

Question 7

An aqueous solution of a non-volatile non-electrolyte has a vapour pressure of 24.5 mm Hg at a temperature where pure water has a vapour pressure of 25.0 mm Hg. What is the approximate freezing point of this solution? (K_f of water = 1.86 K kg mol⁻¹)
  1. −0.4 °C
  2. −2.1 °C
  3. −3.7 °C
  4. −1.1 °C
Show answer
Answer: (B) −2.1 °C
Explanation: The relative lowering is 0.5/25.0 = 0.02, so the solute mole fraction is 0.02 and the water mole fraction is 0.98. That is 0.02/0.98 = 0.0204 mol of solute per mole of water, or per 18 g of water, which gives a molality of about 1.13 mol kg⁻¹. Then ΔT_f = 1.86 × 1.13 ≈ 2.1 K, so the solution freezes at about −2.1 °C.
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Question 8

A dilute aqueous solution contains 3.0 g of a protein (molar mass 60000 g/mol) per litre, the density being close to 1 g/mL. Taking K_f of water = 1.86 K kg/mol, R = 0.083 L bar/(mol K) and T = 300 K for osmotic pressure, which gives the approximate colligative effects and the better method for finding the molar mass?
  1. Freezing point depression ≈ 9.3 × 10⁻⁵ K and osmotic pressure ≈ 1.2 bar; both methods are equally convenient
  2. Freezing point depression ≈ 9.3 × 10⁻³ K and osmotic pressure ≈ 1.2 × 10⁻⁵ bar; depression of freezing point is the better method
  3. Freezing point depression ≈ 9.3 × 10⁻⁵ K and osmotic pressure ≈ 1.2 × 10⁻³ bar; osmotic pressure is the better method
  4. Freezing point depression ≈ 9.3 × 10⁻² K and osmotic pressure ≈ 1.2 bar; depression of freezing point is the better method
Show answer
Answer: (C) Freezing point depression ≈ 9.3 × 10⁻⁵ K and osmotic pressure ≈ 1.2 × 10⁻³ bar; osmotic pressure is the better method
Explanation: The solute concentration is 3.0/60000 = 5.0 × 10⁻⁵ mol per litre, which is nearly the same as the molality. Then the freezing point depression is 1.86 × 5.0 × 10⁻⁵ = 9.3 × 10⁻⁵ K, which is too small to measure reliably. The osmotic pressure is 5.0 × 10⁻⁵ × 0.083 × 300 ≈ 1.2 × 10⁻³ bar, which can be measured at room temperature. This also suits proteins, which are unstable at higher temperatures.

Question 9

Which statement about the desalination of sea water by reverse osmosis is NOT correct?
  1. The pressure applied on the sea water side must be greater than the osmotic pressure of the sea water
  2. Under the applied pressure, water passes through the membrane from the fresh water side into the sea water
  3. The membrane allows water to pass but holds back the dissolved ions and impurities
  4. A thin film of cellulose acetate placed over a suitable support can serve as the membrane
Show answer
Answer: (B) Under the applied pressure, water passes through the membrane from the fresh water side into the sea water
Explanation: In reverse osmosis a pressure larger than the osmotic pressure is applied on the solution side, so pure water is forced out of the sea water through the membrane. This is opposite to the direction of natural osmosis. Cellulose acetate film on a support is a common membrane because it is permeable to water but not to the ions and impurities in sea water.

Question 10

Elements A and B form two non-electrolyte compounds, AB₂ and AB₄. When 1.0 g of AB₂ is dissolved in 20 g of benzene the freezing point falls by 2.0 K, and 1.0 g of AB₄ in 20 g of benzene lowers it by 1.25 K. If Kf of benzene is 5.0 K kg mol⁻¹, the atomic masses of A and B are respectively
  1. 25 and 50
  2. 50 and 37.5
  3. 37.5 and 50
  4. 75 and 25
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Answer: (B) 50 and 37.5
Explanation: Molar mass M = 1000 Kf w/(ΔTf × W) gives AB₂ = 1000 × 5.0 × 1.0/(2.0 × 20) = 125 and AB₄ = 1000 × 5.0 × 1.0/(1.25 × 20) = 200. Subtracting, 2B = 75, so B = 37.5. Then A = 125 − 75 = 50.

Question 11

In a dilute aqueous solution I of a non-volatile non-electrolyte, the mole fraction of solute is 0.04 and the vapour pressure of water is lowered by 1.2 mm Hg. In a second solution II of the same type, at the same temperature, the lowering is 4.2 mm Hg. What is the mole fraction of water in solution II?
  1. 0.86
  2. 0.14
  3. 0.90
  4. 0.96
Show answer
Answer: (A) 0.86
Explanation: The lowering of vapour pressure is proportional to the mole fraction of solute at a given temperature. So x(II) = 0.04 × 4.2/1.2 = 0.14. The mole fraction of water is therefore 1 − 0.14 = 0.86.

Question 12

What mass of a non-volatile, non-electrolyte solute (molar mass 60 g mol⁻¹) must be dissolved in 180 g of water so that the vapour pressure of the solution is exactly 90% of that of pure water? Do not use the dilute solution approximation.
  1. 66.7 g
  2. 54 g
  3. 60 g
  4. 72 g
Show answer
Answer: (A) 66.7 g
Explanation: The relative lowering is 0.10, so the mole fraction of solute must be 0.10. With 10 mol of water, n/(n + 10) = 0.10 gives n = 1.11 mol. The mass is 1.11 × 60 = 66.7 g. The approximation n/10 = 0.10 would give 60 g, which is not accurate at this concentration.

Question 13

Water is to be recovered at 27 °C by reverse osmosis from brine containing 1.17% (mass/volume) NaCl, separated from pure water by a semipermeable membrane. Treating NaCl (molar mass 58.5 g mol⁻¹) as fully dissociated, the pressure applied on the brine side must exceed about (R = 0.082 L atm K⁻¹ mol⁻¹)
  1. 0.98 atm
  2. 19.7 atm
  3. 9.8 atm
  4. 4.9 atm
Show answer
Answer: (C) 9.8 atm
Explanation: Water flows out of the brine only if the applied pressure exceeds its osmotic pressure. A 1.17% solution holds 11.7 g L⁻¹, which is 0.20 M. With i = 2, π = iCRT = 2 × 0.20 × 0.082 × 300 ≈ 9.8 atm.

Question 14

A 0.9% (mass/volume) NaCl solution is isotonic with human red blood cells. What happens when separate samples of the cells are placed in 0.5% and in 1.5% NaCl solutions?
  1. They shrink in 0.5% and swell in 1.5%
  2. They swell in both, because NaCl enters the cells freely
  3. They swell in 0.5% and shrink in 1.5%
  4. They shrink in both, because water leaves the cells whenever NaCl is outside
Show answer
Answer: (C) They swell in 0.5% and shrink in 1.5%
Explanation: The 0.5% solution is hypotonic relative to the cell contents, so water enters by osmosis and the cells swell and may burst. The 1.5% solution is hypertonic, so water moves out and the cells shrink. In each case net water movement across the membrane is towards the higher solute concentration.

Question 15

A certain mass of sucrose (molar mass 342 g mol⁻¹) dissolved in 100 g of water lowers the freezing point by 0.50 K. If the same mass of glucose (molar mass 180 g mol⁻¹) is used in 100 g of water instead, the depression will be about
  1. 0.50 K
  2. 0.26 K
  3. 0.95 K
  4. 1.90 K
Show answer
Answer: (C) 0.95 K
Explanation: Depression of freezing point is proportional to molality. With equal solute mass and equal solvent mass, molality varies inversely with molar mass. Hence ΔTf(glucose) = 0.50 × 342/180 = 0.95 K. Glucose gives more moles per gram, so the depression is larger.

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