Solutions – Class 12 Chemistry NEET MCQs

Solutions – Class 12 Chemistry NEET Practice MCQs

1. The mole fraction of solute in a 1 molal aqueous solution is approximately:
   (a) 0.018
   (b) 0.035
   (c) 0.0177
   (d) 0.27


2. If 0.5 mol of solute is dissolved in 500 g of water, the molality is:
   (a) 0.5 m
   (b) 1.0 m
   (c) 2.0 m
   (d) 0.25 m


3. Which of the following is a colligative property?
   (a) Osmotic pressure
   (b) Surface tension
   (c) Viscosity
   (d) Refractive index


4. The vapour pressure of pure water at 25°C is 23.8 mm Hg. The vapour pressure of a solution containing 18 g of glucose (C₆H₁₂O₆) in 180 g of water will be:
   (a) 23.04 mm Hg
   (b) 22.8 mm Hg
   (c) 23.5 mm Hg
   (d) 23.0 mm Hg


5. Relative lowering of vapour pressure is equal to:
   (a) Mole fraction of solute
   (b) Mole fraction of solvent
   (c) Molality
   (d) Molarity


6. Raoult’s law is applicable to:
   (a) Ideal solutions
   (b) Non-ideal solutions
   (c) Azeotropic mixtures
   (d) Strong electrolytes


7. The elevation in boiling point for 1 molal urea solution is 0.52°C. The boiling point of the solution is:
   (a) 100.52°C
   (b) 99.48°C
   (c) 101°C
   (d) 100°C


8. The depression in freezing point of a 0.1 molal NaCl solution (assuming complete dissociation) is (Kf = 1.86 K kg mol⁻¹):
   (a) 0.186°C
   (b) 0.372°C
   (c) 0.093°C
   (d) 0.279°C


9. The osmotic pressure of a 0.1 M glucose solution at 27°C is (R = 0.0821 L atm K⁻¹ mol⁻¹):
   (a) 2.46 atm
   (b) 1.23 atm
   (c) 4.92 atm
   (d) 0.82 atm


10. Which of the following represents a hypertonic solution?
    (a) Its osmotic pressure is higher than another solution
    (b) Its osmotic pressure is lower
    (c) Same osmotic pressure
    (d) None of these


11. The value of van’t Hoff factor (i) for Na₂SO₄ is:
    (a) 2
    (b) 3
    (c) 1
    (d) 4


12. Which of the following solutions will have the highest boiling point?
    (a) 1 M glucose
    (b) 1 M NaCl
    (c) 1 M CaCl₂
    (d) 1 M urea


13. The osmotic pressure of a 0.01 M NaCl solution at 27°C assuming complete dissociation is approximately:
    (a) 0.24 atm
    (b) 0.48 atm
    (c) 0.12 atm
    (d) 1.0 atm


14. Which of the following pairs forms an ideal solution?
    (a) Benzene and toluene
    (b) Acetone and chloroform
    (c) Ethanol and water
    (d) HCl and water


15. The molarity of 18 g of glucose (C₆H₁₂O₆) in 1 L of solution is:
    (a) 0.5 M
    (b) 0.1 M
    (c) 0.05 M
    (d) 0.2 M


16. If 0.1 mol of BaCl₂ is dissolved in 1 L of water, the total number of ions in the solution will be:
    (a) 0.1 mol
    (b) 0.2 mol
    (c) 0.3 mol
    (d) 0.4 mol


17. Which concentration term is temperature independent?
    (a) Molarity
    (b) Normality
    (c) Molality
    (d) All


18. The vapour pressure of an ideal solution of two volatile liquids A and B is given by:
    (a) PA = PA°xA
    (b) PB = PB°xB
    (c) P = PA°xA + PB°xB
    (d) P = PA°xA × PB°xB


19. For a non-volatile solute, the vapour pressure of the solution compared to pure solvent is:
    (a) Higher
    (b) Lower
    (c) Equal
    (d) Unpredictable


20. The molar mass of a solute that produces a depression of 0.93°C in the freezing point of 100 g water (Kf = 1.86 K kg mol⁻¹) when 5 g is dissolved is:
    (a) 10 g/mol
    (b) 50 g/mol
    (c) 100 g/mol
    (d) 186 g/mol

Answers and Explanations

1. (c) 0.0177 – Mole fraction ≈ 1/(55.5+1)
2. (b) 1.0 m – Molality = 0.5 mol / 0.5 kg
3. (a) Osmotic pressure – depends only on number of solute particles
4. (a) 23.04 mm Hg – P = P₀(1–xsolute)
5. (a) Equal to mole fraction of solute
6. (a) Ideal solutions – obey Raoult’s law
7. (a) 100.52°C – 100 + 0.52
8. (b) 0.372°C – ΔTf = i×Kf×m = 2×1.86×0.1
9. (a) 2.46 atm – π = C R T = 0.1×0.0821×300
10. (a) Osmotic pressure higher → hypertonic
11. (b) 3 – Na₂SO₄ → 2Na⁺ + SO₄²⁻
12. (c) 1 M CaCl₂ – highest i (3)
13. (b) 0.48 atm – π = iCRT = 2×0.01×0.0821×300
14. (a) Benzene and toluene – similar structure and forces
15. (b) 0.1 M – 18/180 = 0.1 mol
16. (c) 0.3 mol – BaCl₂ → Ba²⁺ + 2Cl⁻
17. (c) Molality – based on mass, not volume
18. (c) P = PA°xA + PB°xB
19. (b) Lower – solute reduces vapour pressure
20. (c) 100 g/mol – ΔTf = Kf×m → m = 0.5 mol/kg → molar mass = 5/0.05 = 100