Solutions

12 NEET-UG practice questions on Solutions, part of Chemistry. The first 12 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   Examine these statements about non-ideal solutions. A. A solution showing positive deviation has Δ_mix H > 0 and Δ_mix V > 0. B. A maximum boiling azeotrope forms from a large positive deviation. C. A mixture of chloroform and acetone shows negative deviation from Raoult's law. D. In an ideal solution, A–B interactions are of about the same strength as A–A and B–B interactions. Choose the option listing the correct statements.

   1. AA, C and D only
   2. BA, B and C only
   3. CA and C only
   4. DB and D only

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   Answer: A · A, C and D only

   Weaker A–B attractions give positive deviation with heat absorbed and volume expansion, so A is true. A large positive deviation gives a minimum boiling azeotrope, so B is false. Chloroform hydrogen-bonds to acetone (negative deviation), and D defines an ideal solution.

   • B: Accepts B, which has the azeotrope type reversed, and leaves out D.
   • C: Leaves out D, which is the definition of an ideal solution.
   • D: Accepts B and rejects A and C, which are both true.

   Remember this

   Large positive deviation gives a minimum boiling azeotrope; large negative deviation gives a maximum boiling one.

2. Question 2 · Easy

   A solution is made by dissolving 4.0 g of NaOH in enough water to give 250 mL of solution. What is its molarity? (Molar mass of NaOH = 40 g/mol.)

   1. A16 M
   2. B0.025 M
   3. C0.10 M
   4. D0.40 M

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   Answer: D · 0.40 M

   Moles of NaOH = 4.0/40 = 0.10 mol. Molarity = 0.10/0.250 L = 0.40 M.

   • A: Divides grams by litres, which gives g/L and not mol/L.
   • B: Multiplies moles by the volume instead of dividing.
   • C: Stops at the number of moles and ignores the volume.

   Remember this

   Molarity uses litres of solution, so divide moles by volume in litres and never multiply.

3. Question 3 · Easy

   Gases X and Y are kept over water at the same temperature and the same partial pressure. If the Henry's law constant of X is larger than that of Y, then X is

   1. Ainsoluble in water at this pressure
   2. Bequally soluble in water as Y
   3. Cless soluble in water than Y
   4. Dmore soluble in water than Y

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   Answer: C · less soluble in water than Y

   Henry's law is p = K_H x, so x = p/K_H. At a fixed p, a larger K_H means a smaller mole fraction of the gas in solution.

   • A: A larger K_H lowers the solubility but does not make it zero.
   • B: Ignores K_H, which differs between the gases.
   • D: Reads K_H as a direct measure of solubility; it varies inversely.

   Remember this

   Higher K_H means lower solubility; K_H also rises with temperature, so gases dissolve less on warming.

4. Question 4 · Medium

   Find the boiling point of a solution of 5.85 g of NaCl in 500 g of water, assuming complete dissociation. (NaCl = 58.5 g/mol; K_b of water = 0.52 K kg/mol; boiling point of water = 100.000 °C.)

   1. A100.208 °C
   2. B100.104 °C
   3. C100.052 °C
   4. D100.416 °C

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   Answer: A · 100.208 °C

   Moles of NaCl = 0.10, so m = 0.10/0.500 = 0.20 m. With i = 2, ΔT_b = 2 × 0.52 × 0.20 = 0.208 K, which gives 100.208 °C.

   • B: Ignores the van't Hoff factor and takes i = 1.
   • C: Uses 0.10 mol as the molality and also takes i = 1.
   • D: Doubles the molality as well as applying i = 2.

   Remember this

   For an electrolyte use ΔT_b = i K_b m; NaCl has i = 2 when fully dissociated.

5. Question 5 · Medium

   An aqueous NaOH solution is 20% by mass. What is its molality? (Molar mass of NaOH = 40 g/mol.)

   1. A0.625 m
   2. B4.00 m
   3. C6.25 m
   4. D5.00 m

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   Answer: C · 6.25 m

   In 100 g of solution there are 20 g NaOH (0.50 mol) and 80 g water. Molality = 0.50/0.080 kg = 6.25 m.

   • A: Slips a factor of ten in converting grams to kilograms.
   • B: Takes the ratio of the two masses, 80/20, in place of moles per kilogram.
   • D: Divides by 100 g of solution instead of 80 g of solvent.

   Remember this

   Molality divides by the mass of solvent alone, so subtract the solute from the solution first.

6. Question 6 · Medium

   A solution containing 2.0 g of a protein in 200 mL of solution has an osmotic pressure of 0.0083 bar at 27 °C. Determine the molar mass of the protein. (Take R = 0.083 L bar K⁻¹ mol⁻¹.)

   1. A2700 g/mol
   2. B30000 g/mol
   3. C6000 g/mol
   4. D150000 g/mol

   Answer and explanation in LearnFree account

7. Question 7 · Easy

   Arrange these aqueous solutions in increasing order of freezing point, assuming complete dissociation of the salts. A. 0.1 m KCl B. 0.1 m glucose C. 0.1 m Al₂(SO₄)₃ D. 0.1 m K₂SO₄

   1. AB, A, D, C
   2. BD, C, A, B
   3. CC, D, A, B
   4. DC, A, D, B

   Answer and explanation in LearnFree account

8. Question 8 · Medium

   Match List-I with List-II for dilute aqueous solutions, assuming complete dissociation of the ionic compounds. List-I (solute) A. Urea B. MgSO₄ C. CaCl₂ D. K₄[Fe(CN)₆] List-II (van't Hoff factor) I. 3 II. 1 III. 5 IV. 2

   1. AA-II, B-IV, C-III, D-I
   2. BA-II, B-I, C-IV, D-III
   3. CA-II, B-IV, C-I, D-III
   4. DA-IV, B-II, C-I, D-III

   Answer and explanation in LearnFree account

9. Question 9 · Medium

   Assertion A: Aquatic species are more comfortable in cold water than in warm water. Reason R: The solubility of oxygen in water increases with a rise in temperature. Choose the correct answer.

   1. ABoth A and R are correct but R is not the correct explanation of A
   2. BBoth A and R are correct and R is the correct explanation of A
   3. CA is correct but R is not correct
   4. DA is not correct but R is correct

   Answer and explanation in LearnFree account

10. Question 10 · Medium

    Water has a vapour pressure of 760 mmHg at 100 °C. If 18 g of glucose is dissolved in 178.2 g of water, what is the vapour pressure of the solution at this temperature? (Glucose = 180 g/mol, water = 18 g/mol.)

    1. A744.8 mmHg
    2. B7.6 mmHg
    3. C752.4 mmHg
    4. D684.0 mmHg

    Answer and explanation in LearnFree account

11. Question 11 · Hard

    Benzoic acid dimerises in benzene. A solution of 6.1 g of benzoic acid (M = 122 g/mol) in 50 g of benzene shows a freezing point depression of 4.0 K. Taking K_f of benzene as 5.0 K kg/mol, what percentage of the acid is dimerised?

    1. A20%
    2. B40%
    3. C60%
    4. D80%

    Answer and explanation in LearnFree account

12. Question 12 · Medium

    Liquids A and B form an ideal solution. At a given temperature, their pure vapour pressures are 120 mmHg and 40 mmHg. For a liquid mixture with mole fraction of A = 0.25, the mole fraction of A in the vapour is

    1. A0.33
    2. B0.50
    3. C0.25
    4. D0.75

    Answer and explanation in LearnFree account