Correct answer. A non-volatile solute lowers the freezing point of a solvent. Colligative properties depend upon the number of particles of the solute and is independent of nature of the solute.

Correct answer. Elevation in boiling point (ΔT_b) = K_b x iNaCl x mNaCl = (0.51^oC kg mol^-1) x (2) x (0.25 mol/kg) = 0.26^oC The boiling point of 0.25 molal NaCl = 100.00^oC + 0.26^oC = 100.26^oC

Correct answer. Urea is non-volatile. Water obeys Raoult’s law. Pwater = Xwater x P°water 266.3 torr = Xwater x (355.1 torr) Xwater = 266.3 / 355.1 = 0.75 From equation, X₁ + X₂ + X₃ ...... = 1 Xurea + Xwater = 1 Xurea = 1 – 0.75 = 0.25

Correct answer. Among the given options, ammonia has the highest solubility in water. 

Correct answer. Similar molalities, so similar deviations from expected. Expected van't Hoff factor of solutions at given concentration: i expected for BaCl₂ = 0.10(3) = 0.30 i expected for K₂SO₄ = 0.12(3) = 0.36 i expected for KBr = 0.12(2) = 0.24 Higher the van’t Hoff factor, higher will be the boiling point. Hence, correct order of boiling points is KBr < BaCl₂ < K₂SO₄.

Correct answer. 100.0 g of 45% mix: 45.0 g C₂H₂(OH)₂ + 55.0 g H₂O nglycol = 45.0 g / 62.0 g mol^-1 = 0.73 mol mglycol = (0.73 mol / 0.055 kg) = 13.272 mol Kg^-1 ΔT_f = (1.86^oC kg mol^-1) x (13.272 mol Kg^-1) = 24.7^oC Freezing point = 0.00°C – 24.7°C = -24.7^oC

Correct answer. The amount of KOH required is (0.120 L) × (0.10 mol L⁻¹) = 0.012 mol. The molar mass of KOH is 56.1 g. So, the weight of KOH required = 0.012 mol × 56.1 gmol⁻¹ = 0.67 g.

Correct answer. 100 g of solution contains 65/342 = 0.19 mol of sugar and 36/18 = 2 mol of water  The mole fraction of water in the solution = 2 / (0.19 + 2) = 0.91 So, its vapour pressure will be 0.91 × 26.0 torr = 23.7 torr

Correct answer. Number of moles of solute per litre of the solution is molarity. Molality is better concentration term than molarity.

Correct answer.

Number of moles of ethanol in 670.3 g = 670.3 g/ 46 g mol^-1 = 14.6 mol Molarity = Number of moles of solute / Volume of solution (in L) = 14.6 mol / 1 L = 14.6 mol L^-1

Correct option: Both enthalpy change and volume of mixing change should be zero for an ideal solution.

Correct answer. Assume that we use 1 L of ethylene glycol and 3 L of water. The mass of ethylene glycol will be 1.11 kg and that of water will be 3.0 kg. So, the total mass of the solution is 4.11 kg. Number of moles of ethylene glycol = 1110 g / 62 g mol^-1 = 17.9 mol Molality of ethylene glycol = 17.9 mol / 4.11 Kg = 4.36 mol Kg^-1 Freezing point depression ΔT_f = K_f x molality of solution For water, K_f = 1.86 K kg⁻¹ mol ΔT_f= (1.86 K kg⁻¹ mol) x (4.36 mol Kg^-1) = 8.1 K

Correct answer. As bonding between two components (acetone and ethyl alcohol) is more than individuals due to H-bonding, this mixture will show negative deviation from Raoult’s law.

Correct answer. More the van’t Hoff factor, more is the boiling point.

Correct answer. Colligative properties are properties of solutions that depend upon the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the type of chemical species present.

Correct relation for vapour pressure of pure solvent (P⁰) and vapour pressure of solution (P) is P⁰ > P.

Correct answer. Depression in freezing point is a colligative property as it depends on number of solute particles. Elevation in boiling point is a colligative property as it depends on number of solute particles. Osmotic pressure is a colligative property as it depends on number of solute particles.

Correct option

Correct answer is A – 3; B – 2; C – 1.

Pressure is directly proportional to moles of gases.  4 grams of He = 1 mole of He gas (weight/molar weight); highest among all. Molecular weight of Helium gas is 4.