5: Colligative Properties

Correct answer: C. एक से अधिक

Explanation: The van’t Hoff factor i represents the ratio of the actual number of solute particles in solution to the number expected without association or dissociation. Ionisation splits a solute into two or more ions, increasing the effective particle count. Since ΔT_f = iK_fm, a larger-than-expected depression means i > 1.

Correct answer: B. संघटन

Explanation: For a dilute solution, boiling-point elevation is given by ΔT_b = iK_bm. Association causes two or more solute particles to combine into larger units, so the number of independent particles decreases and i becomes less than one. Consequently, the observed ΔT_b is smaller than the value calculated for an undissociated solute.

Correct answer: B. ऐसे विलेय का विलयन जो दो आयन देता है

Explanation: Boiling-point elevation follows ΔT_b = iK_bm. At the same molality and with the same solvent, K_b and m are fixed, so the elevation depends on the van’t Hoff factor i. A non-electrolyte gives i approximately 1, whereas a solute that completely forms two ions gives i approximately 2. Therefore option B produces the greater elevation.

Correct answer: A. बहुत तनु विलयन में भी परासरण दाब मापा जा सकता है

Explanation: Large molecules such as proteins and polymers may decompose or undergo structural changes when their solutions are heated, so boiling-point or freezing-point methods can be unsuitable. Osmotic pressure can be measured accurately at or near room temperature using a very dilute solution. Since π = CRT, the method gives molar mass without requiring a large amount of solute or heating.

Correct answer: B. कोशिका से बाहर

Explanation: A hypertonic external solution has a higher effective solute concentration, and therefore a lower water potential, than the cell interior. Across a selectively permeable membrane, water moves by osmosis from the side with higher water potential to the side with lower water potential. Thus water leaves the cell, causing it to shrink or become plasmolysed.

Correct answer: A. विलायक का कोई शुद्ध प्रवाह नहीं होगा

Explanation: Solutions having equal osmotic pressure at the same temperature are isotonic with respect to the membrane. Although individual solvent molecules may cross the semipermeable membrane in both directions, the two opposing rates are equal. Therefore there is no net movement of solvent from one solution to the other, and no volume change due to osmosis occurs.

Correct answer: A. क्योंकि यह संबंध निरपेक्ष तापमान पर आधारित है

Explanation: The equation π = iCRT is derived from the ideal-solution or gas-like relationship in which temperature is measured from absolute zero. The kelvin scale is an absolute scale, so its zero corresponds to the minimum theoretical thermal energy and ratios of temperature remain physically meaningful. Using Celsius directly would give an incorrect proportionality because 0 °C is not absolute zero.

Correct answer: B. The lowering increases

Explanation: Lowering of vapour pressure is a colligative property, so it depends on the effective number of solute particles rather than their chemical identity. When more particles are present, fewer solvent molecules can escape into the vapour phase. Consequently, the vapour pressure decreases by a larger amount. Ionisation increases the particle count and therefore can increase the lowering.

Correct answer: A. Because the vapour pressure of the solution decreases

Explanation: A non-volatile solute does not contribute appreciably to the vapour phase, but it reduces the escaping tendency of solvent molecules. Thus, at any given temperature, the solution has a lower vapour pressure than the pure solvent. The solution must be heated to a higher temperature before its vapour pressure equals the external pressure, so its boiling point rises.

Correct answer: A. Because the solute interferes with the orderly freezing of solvent molecules

Explanation: Dissolved solute particles disturb the formation of an ordered crystal lattice by solvent molecules. Therefore, the chemical potential of the liquid solvent becomes relatively lower than that of the pure solvent, and a lower temperature is required for solid-liquid equilibrium. This decrease in freezing temperature is called depression of freezing point and is a colligative effect.

Correct answer: B. 0.90 K

Explanation: First convert the solvent mass into kilograms because molality is moles of solute per kilogram of solvent: 200 g = 0.200 kg. Therefore, m = 0.1/0.200 = 0.5 mol kg⁻¹. For a non-electrolyte, i = 1, so ΔT_f = iK_fm = 1 × 1.8 × 0.5 = 0.90 K. Thus, option B is correct.

Correct answer: B. 0.25 K

Explanation: Convert the solvent mass first: 400 g equals 0.400 kg. The molality is therefore m = 0.2/0.400 = 0.5 mol kg⁻¹. Since the solute is a non-electrolyte, the van’t Hoff factor is i = 1. Applying ΔT_b = iK_bm gives ΔT_b = 1 × 0.5 × 0.5 = 0.25 K. Therefore, the correct answer is option B.

Correct answer: B. They depend on the effective number of solute particles

Explanation: Colligative properties are properties of dilute solutions that depend primarily on the number of dissolved particles relative to the amount of solvent, not on the chemical identity, colour, or shape of the container. Their magnitude is affected by dissociation or association because these processes change the effective particle count. Thus, option B is the correct statement.

Correct answer: C. 3

Explanation: The van’t Hoff factor i is the ratio of the actual number of solute particles in solution to the number expected if no dissociation or association occurred. Under ideal complete dissociation, one formula unit producing three ions gives three effective particles. Therefore, i = 3, and option C is correct. This value increases colligative effects threefold compared with an undissociated solute.

Correct answer: A. They will be the same

Explanation: Freezing-point depression is given by ΔT_f = iK_fm. For non-electrolytes under ideal conditions, i = 1. Because both solutions use the same solvent, they have the same K_f, and the stated molalities are equal. Every factor in the equation is therefore identical, so both solutions have the same ΔT_f. Option A is correct.

Correct answer: A. The one in which ionisation occurs

Explanation: Colligative properties depend on the total number of dissolved particles, not merely on the original number of solute formula units. Ionisation changes one solute unit into two or more ions, increasing the effective particle concentration. Therefore, at the same molality, the ionising solute produces the greater elevation, depression, or osmotic effect. This increase is represented by the van’t Hoff factor, i, which is greater than one for ideal dissociation.

Correct answer: A. Identify the applicable property and formula

Explanation: A colligative-properties problem must first be classified correctly. Determine whether it involves relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, or osmotic pressure. Then select the corresponding equation, identify quantities such as molality, mole fraction, concentration, or van’t Hoff factor, convert units consistently, and only afterward substitute values and calculate.

Correct answer: A. Their molar concentrations are equal

Explanation: For an ideal dilute solution, osmotic pressure is given by π = CRT, where C is the molar concentration, R is the gas constant, and T is absolute temperature. Because both solutions have the same π and the same T, and R is universal, their molar concentrations must be equal. This does not require identical solutes, colours, or densities; those properties depend on chemical identity and composition.

Correct answer: B. Because the mole fraction and escaping tendency of solvent molecules decrease

Explanation: When a non-volatile solute is dissolved, it contributes essentially no vapour above the solution. It reduces the mole fraction of the solvent and therefore reduces the fraction of surface molecules able to escape into the vapour phase. Raoult’s law expresses this as p₁ = x₁p₁⁰; because x₁ is less than one, the solvent vapour pressure is lower than that of the pure solvent at the same temperature.

Correct answer: D. Al₂(SO₄)₃

Explanation: The depression in freezing point is given by ΔT_f = iK_fm. Since all solutions have the same solvent and the same molality, K_f and m are constant, so the largest van’t Hoff factor produces the greatest depression. Glucose gives one particle, NaCl gives two ions, CaCl₂ gives three ions, and Al₂(SO₄)₃ gives five ions on complete dissociation. Therefore option D is correct.

Correct answer: B. Partial dissociation has occurred

Explanation: KCl ideally dissociates as KCl → K⁺ + Cl⁻, so one formula unit can produce two particles and the ideal van’t Hoff factor is 2. A measured value of 1.8 lies between the value for no dissociation, approximately 1, and the value for complete dissociation, 2. Thus the particles are not completely separated; the result indicates partial dissociation, possibly with some ion pairing.

Correct answer: C. CaCl₂

Explanation: Freezing-point depression is calculated using ΔT_f = iK_fm. At equal molality in the same solvent, the solution with the largest van’t Hoff factor has the greatest depression and therefore the lowest actual freezing point. Glucose and urea are non-electrolytes with i approximately 1, NaCl gives about 2 ions, and CaCl₂ gives about 3 ions. Hence CaCl₂ is correct.

Correct answer: B. 0.02

Explanation: For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p⁰ − p)/p⁰ = x₂, where x₂ is the solute mole fraction. Therefore, if Δp/p⁰ is 0.02, the approximate solute mole fraction is x₂ = 0.02. The dilute-solution assumption supports this simple relation and rules out values such as 0.20 or 2.0.

Correct answer: C. 0.744 K

Explanation: The relation for freezing-point depression is ΔT_f = iK_fm. Glucose is a non-electrolyte, so i = 1, and its given depression is 0.372 K. At the same molality and in the same solvent, completely dissociated NaCl produces Na⁺ and Cl⁻, so i = 2. Its depression is therefore twice the glucose value: 2 × 0.372 = 0.744 K. Hence option C is correct.

Correct answer: B. Dissociation of solute

Explanation: The elevation in boiling point is given by ΔTb = iKb m. If the expected calculation assumes i = 1 but the solute dissociates into two or more particles, the van’t Hoff factor becomes greater than 1. Consequently, the number of solute particles increases and the observed boiling-point elevation becomes larger than the expected value. Therefore, dissociation is the correct reason.