5: Colligative Properties

Correct answer: C. Change in surface tension

Explanation: Relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure are colligative properties. For a given concentration, they depend primarily on the number of dissolved particles. Surface tension is not colligative; it depends on the nature of the solute and solvent and on intermolecular forces. Therefore, change in surface tension is correct.

Correct answer: C. Osmotic pressure

Explanation: Osmotic pressure is a colligative property. In a dilute solution it is represented by π = iCRT, so it is governed by the effective number or concentration of solute particles, with i accounting for dissociation or association. Viscosity, surface tension and refractive index depend substantially on the chemical nature and intermolecular interactions of the substances. Hence option C is correct.

Correct answer: B. NaCl solution

Explanation: Freezing-point depression is given by ΔTf = iKf m. Glucose is a non-electrolyte and remains as molecules, so its van’t Hoff factor is approximately 1. Completely dissociated NaCl produces Na+ and Cl−, giving i approximately 2. At equal molality, NaCl therefore has more effective solute particles and a greater freezing-point depression.

Correct answer: B. Association of solute particles

Explanation: When solute molecules associate, two or more individual particles combine to form one larger particle. The total number of solute particles therefore becomes smaller than expected. Since colligative properties depend on the number of particles, the observed effect is reduced. If molar mass is calculated from this smaller effect without correcting for association, the calculated or observed molar mass becomes greater than the actual molar mass. Association corresponds to a van’t Hoff factor less than one.

Correct answer: C. CaCl₂

Explanation: The elevation in boiling point is ΔTb = iKb m. Since all solutions have the same molality and solvent, Kb and m are common; therefore, the solution with the largest van’t Hoff factor has the greatest elevation. Glucose and urea do not dissociate, so i = 1. NaCl gives two ions, i = 2, whereas CaCl₂ gives three ions, i = 3, on complete dissociation. Thus CaCl₂ has the highest boiling point.

Correct answer: A. Because it does not depend on volume change with temperature

Explanation: Molality is defined as the number of moles of solute per kilogram of solvent, so it is based on mass. Mass remains essentially constant when temperature changes, whereas the volume of a solution can expand or contract. Since molarity uses solution volume, it may vary with temperature. Therefore, molality gives more reliable colligative-property calculations.

Correct answer: C. 0.3

Explanation: For a solution in which dissociation or association is represented by the van’t Hoff factor, the boiling-point elevation is calculated using ΔT_b = iK_bm. Substituting the given values gives ΔT_b = 3 × 0.5 × 0.2 = 0.3. Therefore, the boiling point is elevated by 0.3 temperature units, so option C is correct.

Correct answer: B. Association

Explanation: The van’t Hoff factor compares the actual number of solute particles with the number expected without association or dissociation. When i is less than 1, solute particles combine to form larger associated units, reducing the number of effective particles. Since i = 0.5, the solution shows association, so option B is correct.

Correct answer: A. The mole fraction of the solvent

Explanation: For a solution containing a non-volatile solute, Raoult’s law states that the partial vapour pressure of the solvent is p_A = x_A p_A°, where x_A is the mole fraction of the solvent and p_A° is the vapour pressure of the pure solvent. Thus, at a fixed temperature, the partial pressure is proportional to the solvent’s mole fraction.

Correct answer: A. The lowering increases

Explanation: For a dilute solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as approximately equal to the mole fraction of the solute, x₂. Therefore, when x₂ increases, the fraction of solvent molecules escaping into the vapour phase decreases, the solution vapour pressure falls further, and the lowering of vapour pressure increases. Option A is correct.

Correct answer: A. Osmotic pressure is measurable even in very dilute solutions

Explanation: Large molecules such as proteins and polymers may decompose when heated or may not form a stable vapour, so boiling-point or vapour-pressure methods are unsuitable. Their solutions can, however, be prepared in very dilute form, and osmotic pressure can still be measured at ordinary temperature. Since π = CRT, the measured pressure provides molar-mass information without heating or vaporising the macromolecule. Thus A is correct.

Correct answer: A. The number of effective solute particles decreases

Explanation: Freezing-point depression is proportional to the effective number of solute particles, expressed as ΔTf = iKf m. During association, two or more solute particles combine to form one larger species, so the van’t Hoff factor i becomes less than the value expected for independent particles. The effective particle concentration therefore falls, and the observed depression is smaller. Hence A is correct.

Correct answer: B. Nearly double

Explanation: One formula unit of AB produces two independently moving ions, A⁺ and B⁻, on complete dissociation. Thus the van’t Hoff factor is approximately i = 2, while a non-dissociating solute has i = 1. Since colligative effects are proportional to i at the same molality and temperature, the dissociated solute produces nearly twice the effect. Therefore, B is correct.

Correct answer: A. (0.1)

Explanation: The relative lowering of vapour pressure is calculated using (p° − p)/p°, where p° is the vapour pressure of the pure solvent and p is the vapour pressure of the solvent in the solution. Here, the lowering is 80 − 72 = 8. Dividing this lowering by the pure-solvent vapour pressure gives 8/80 = 0.10. Thus, the relative lowering is dimensionless and option A is correct.

Correct answer: A. The solute can also contribute to vapour pressure

Explanation: The simple vapour-pressure-lowering expression assumes that the solute is non-volatile, so the vapour phase is produced almost entirely by the solvent. A volatile solute also evaporates and contributes its own partial pressure. The total vapour pressure must then be described using the partial pressures of both components, usually through Raoult’s law, rather than by the simplified relation alone.

Correct answer: A. They will be equal

Explanation: For a dilute solution, osmotic pressure is π = iCRT. A non-dissociating solute has i = 1, and at the same temperature R and T are common to both solutions. Therefore, equal osmotic pressures require equal molar concentrations: C1RT = C2RT, so C1 = C2. Such solutions are called isotonic with respect to osmotic pressure.

Correct answer: B. Mole fraction of solute

Explanation: For a solution containing a non-volatile solute, only the solvent contributes appreciably to the vapour pressure. According to Raoult’s law, the relative lowering of vapour pressure is (p° − p) / p° = Xsolute, where p° is the vapour pressure of the pure solvent and p is that of the solution. Therefore, the correct answer is the mole fraction of the solute, option B. This is a colligative property because it depends on the number of solute particles, not their chemical identity.

Correct answer: A. 0.25

Explanation: The mole fraction of a component is calculated by dividing its number of moles by the total number of moles of all components present. Here, total moles = moles of solute + moles of solvent = 0.25 + 0.75 = 1.00. Therefore, the mole fraction of the solute is Xsolute = 0.25 / 1.00 = 0.25. Hence, option A is correct. The mole fractions of all components together must add up to 1.

Correct answer: A. \(1.5^{\circ}C\)

Explanation: Depression in freezing point means that the freezing point of a solution is lower than that of the pure solvent. The relation is \(\Delta T_f=T_f^0-T_f\), where \(T_f^0\) is the freezing point of the pure solvent and \(T_f\) is that of the solution. Therefore, \(T_f=4.0-2.5=1.5^{\circ}C\). Hence, option A is correct.

Correct answer: C. Calcium chloride solution

Explanation: For equal molality, freezing-point depression is given by \(\Delta T_f=iK_fm\). Since the solvent, \(K_f\), and molality are the same, the result depends on the van’t Hoff factor \(i\), or the effective number of particles. Urea and glucose have \(i\approx1\), NaCl gives about two ions, and CaCl₂ can give three ions ideally. Thus CaCl₂ produces the greatest depression, so C is correct.

Correct answer: A. Ionisation of solute

Explanation: Ionisation breaks one solute formula unit into two or more particles, increasing the effective number of particles in solution. Consequently, the observed colligative property is greater than expected for the same amount of solute. When molar mass is calculated from this enlarged effect, the calculated or apparent molar mass becomes smaller than the true value. Therefore, ionisation is the correct reason.

Correct answer: B. Association

Explanation: The van’t Hoff factor \(i\) compares the actual number of particles in solution with the number expected without association or dissociation. If \(i<1\), the actual particle count is smaller than expected. Association causes two or more solute molecules to combine into one larger species, reducing particle number. Therefore, an \(i\) value below unity most commonly indicates association, making B correct.

Correct answer: B. \(0.2\)

Explanation: For a dilute solution, osmotic pressure is related to molar concentration by the van’t Hoff equation \(\pi=CRT\). Rearranging gives \(C=\pi/(RT)\). Substituting the values, \(C=4.92/(0.082\times300)=4.92/24.6=0.20\). Therefore, the molar concentration is \(0.2\) in the consistent concentration unit, and option B is correct.

Correct answer: B. \(1.23\)

Explanation: The osmotic pressure of a dilute solution is calculated using \(\pi=CRT\), where \(C\) is molar concentration, \(R\) is the gas constant, and \(T\) is absolute temperature. Substitution gives \(\pi=0.05\times0.082\times300\). Since \(0.082\times300=24.6\), the product is \(0.05\times24.6=1.23\). Thus option B is correct.

Correct answer: B. When association occurs

Explanation: Association occurs when two or more solute molecules combine to form a single larger species. This decreases the number of particles in solution and therefore reduces the observed colligative effect. If molar mass is calculated from this smaller-than-expected effect, the result is an apparent molar mass greater than the actual molar mass. Hence association is the correct answer, whereas ionisation would usually lower the apparent value.