5: Colligative Properties

The elevation in boiling point is given by \(\Delta T_b=iK_bm\), where \(i\) represents the effective number of particles and \(m\) is molality. Since the solvent and its ebullioscopic constant \(K_b\) are unchanged, a larger \(\Delta T_b\) indicates a larger effective particle concentration, caused by more solute particles or dissociation. Therefore, option A is correct.

The depression in freezing point is given by \(\Delta T_f=iK_fm\), where \(i\) is the van’t Hoff factor and \(m\) is molality. Since \(K_f\) is identical in every option, compare only the product \(i\times m\). The products are 1, 1, 3 and 2 respectively. The greatest value is 3, obtained in option C; therefore option C gives the maximum freezing-point depression.

For elevation of boiling point, the relation is ΔT_b = iK_bm. A non-electrolyte neither dissociates nor associates appreciably in solution, so its van’t Hoff factor is i = 1. Substituting the given values gives ΔT_b = 1 × 0.52 × 0.5 = 0.26 K. Therefore, the boiling point rises by 0.26 K, so option A is correct.

The depression in freezing point for a solution is calculated using ΔT_f = iK_fm. Here, i = 0.5 accounts for the effective number of solute particles, K_f = 2 K kg mol⁻¹, and m = 1 mol kg⁻¹. Thus, ΔT_f = 0.5 × 2 × 1 = 1.0 K. Hence, the freezing point decreases by 1.0 K and option B is correct.

Freezing-point depression is given by ΔTf = iKf m. Since all solutions have the same molality and the same solvent, Kf and m are constant, so the solution with the largest van’t Hoff factor has the greatest depression and therefore the lowest freezing point. Glucose gives i = 1, NaCl gives 2, CaCl₂ gives 3, and Al₂(SO₄)₃ gives 5 ions on complete dissociation. Hence option D is correct.

For a dilute solution, solute particles are sufficiently far apart, so their mutual interactions are small and the solution behaves nearly ideally. The relation π = CRT, used to calculate molar mass from osmotic pressure, is most reliable under this condition. In concentrated solutions, association, dissociation, and intermolecular forces can cause deviations and produce an inaccurate molar mass.

NaCl dissociates according to NaCl → Na⁺ + Cl⁻, so one formula unit produces two particles when completely dissociated. For a substance producing n particles, the van't Hoff factor is i = 1 + α(n − 1), where α is the degree of dissociation. Here α = 60% = 0.60 and n = 2. Therefore, i = 1 + 0.60(2 − 1) = 1.60. Thus, option C is correct.

For dissociation of one formula unit into two ions, the van’t Hoff factor is i = 1 + α, where α is the degree of dissociation. Here, 50% dissociation means α = 0.50. Therefore, i = 1 + 0.50 = 1.50. Thus, the solution has 1.5 times as many solute particles as the corresponding non-dissociated solution, and its colligative effect is increased by the same factor.

For freezing-point depression, the relation is ΔTf = iKf m. A non-dissociating solute has van’t Hoff factor i = 1. Ideally, NaCl completely dissociates into Na⁺ and Cl⁻ ions, so i = 2. Since both solutions have the same molality and solvent, the NaCl solution produces twice the number of solute particles and therefore approximately twice the freezing-point depression. This is an ideal calculation; real ion interactions may cause a small deviation.

One formula unit of aluminium sulfate, Al₂(SO₄)₃, dissociates completely as Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. Thus, one formula unit produces a total of five ions in solution. For complete dissociation, the ideal van’t Hoff factor equals the number of particles formed, so i = 2 + 3 = 5. Therefore, option C is correct.

For an electrolyte solution, osmotic pressure is calculated using \(\pi=iCRT\). Complete dissociation of one formula unit of \(\mathrm{CaCl_2}\) produces one \(\mathrm{Ca^{2+}}\) ion and two \(\mathrm{Cl^-}\) ions, so the van’t Hoff factor is \(i=3\). Therefore, \(\pi=3\times0.05\times0.082\times300=3.69\,\mathrm{atm}\). Hence, option C is correct. Omitting the factor 3 would incorrectly treat the electrolyte as a nonelectrolyte.

On complete dissociation, \(\mathrm{Ca(NO_3)_2}\) forms three ions: one \(\mathrm{Ca^{2+}}\) and two \(\mathrm{NO_3^-}\), so \(n=3\). For dissociation, the relation is \(i=1+\alpha(n-1)\). Substituting the values gives \(2.6=1+\alpha(3-1)=1+2\alpha\), so \(\alpha=0.8\). Converting the fraction to a percentage gives \(0.8\times100=80\%\). Therefore, option C is correct.

The governing relation for freezing-point depression is ΔTf = iKf m. A non-dissociated solute has van’t Hoff factor i = 1. The given depression is ΔTf = 1.86 K, so m = ΔTf/(iKf) = 1.86/(1 × 1.86) = 1.0 mol kg−1. The other choices do not satisfy the equation with the stated Kf and depression. Hence, option B is correct.

Osmotic pressure is a colligative property, so it depends on the total number of dissolved particles rather than their chemical identity. For a dilute solution, π = iCRT, where i accounts for dissociation or association. It also depends on temperature, while a pure solvent has π = 0. Dissociation generally increases particle number and osmotic pressure.

When a solute dissociates, one formula unit produces two or more particles in the solution. Colligative properties depend on the total number of solute particles, so the observed effect becomes larger than expected for the undissociated solute. If the usual formula is applied without correcting for dissociation, the calculated number of moles becomes too large and the observed molar mass appears smaller than the true molar mass. This corresponds to a van’t Hoff factor greater than one, i > 1.

Ethanoic acid molecules form hydrogen-bonded dimers in a non-polar solvent such as benzene. Thus, two solute molecules behave approximately as one dissolved particle. The number of effective particles decreases, so the observed colligative effect is smaller and the van’t Hoff factor becomes less than one. Sodium chloride and potassium sulfate generally dissociate in water, giving i greater than one, while glucose remains essentially undissociated and has i close to one.

For AB → A⁺ + B⁻, one original particle produces two particles on complete dissociation. If the degree of dissociation is α, the van’t Hoff factor is i = 1 + α(ν − 1), where ν = 2 is the total number of particles formed. Here α = 70/100 = 0.70, so i = 1 + 0.70(2 − 1) = 1.70. Therefore, option C is correct.

The van’t Hoff factor compares the actual number of solute particles in solution with the number expected from the undissociated formula units. A value greater than 1 means that the particle count has increased. This occurs when the solute dissociates or ionises into two or more particles, so dissociation is the correct explanation.

For isotonic solutions at the same temperature, osmotic pressures are equal. Since π = iCRT, the products iC must be equal. Glucose does not dissociate, so i = 1 and iC = 1 × 0.3 = 0.3. Complete dissociation of CaCl₂ gives Ca²⁺ and two Cl⁻ ions, so i = 3. Therefore, 3C = 0.3 and C = 0.10 M. Hence, option B is correct.

Na₂SO₄ dissociates as Na₂SO₄ → 2Na⁺ + SO₄²⁻, producing three particles when completely dissociated. For a degree of dissociation α = 0.50, the van’t Hoff factor is i = 1 + α(n − 1) = 1 + 0.50(3 − 1) = 2. The effective molality is therefore i × m = 2 × 0.20 = 0.40 mol kg⁻¹. Thus, option C is correct.

For aqueous solutions, ΔT_f = iK_fm. Since the solvent and molality are the same, compare only the van’t Hoff factor i. Glucose remains as one particle, NaCl forms two ions, BaCl₂ forms three ions, and Al₂(SO₄)₃ forms two Al³⁺ ions plus three sulfate ions, giving five particles. Thus Al₂(SO₄)₃ has the largest i and the greatest freezing-point depression.

Boiling-point elevation follows ΔT_b = iK_bm. The reference non-dissociated solute has i = 1 and gives ΔT_b = 0.052 K at 0.1 m. Complete dissociation of K₂SO₄ produces 2K⁺ and SO₄²⁻, so there are three particles and i = 3. At the same molality and solvent, the elevation becomes three times the reference value: 3 × 0.052 = 0.156 K. Therefore, option C is correct.

The relation for freezing-point depression is ΔT_f = iK_fm. With the same solvent and equal molality, the largest value of i gives the greatest depression. Glucose gives one particle, NaCl gives two ions, CaCl₂ gives three ions, and Al₂(SO₄)₃ dissociates into two Al³⁺ ions and three SO₄²⁻ ions, giving five particles. Therefore, Al₂(SO₄)₃ produces the greatest depression and option D is correct.

Freezing-point depression is given by ΔT_f = iK_fm. Because K_f and molality are the same for all four solutions, the solution with the largest van’t Hoff factor i gives the greatest depression. Urea gives one particle, KCl gives two, CaCl₂ gives three, and Al₂(SO₄)₃ gives five ions: 2Al³⁺ and 3SO₄²⁻. Therefore, option D is correct.

Osmotic pressure depends on the total number of dissolved particles and is given by π = iCRT. Glucose does not ionise, so i = 1 and its effective particle concentration is 0.1 M. Completely dissociated NaCl gives Na⁺ and Cl⁻, so i = 2 and its effective concentration is 0.2 M. Water therefore flows toward the NaCl side, which has the greater osmotic pressure.