5: Colligative Properties

Osmotic pressure follows π = iCRT. Urea is a non-electrolyte, so its van’t Hoff factor is 1. Equating the pressures at the same temperature gives i_unknown × 0.05 = 1 × 0.10. Thus i_unknown = 0.10/0.05 = 2. The unknown solute therefore produces twice as many effective particles per formula unit as a non-electrolyte.

The van’t Hoff factor measures the ratio of actual dissolved particles to the particles expected without association or dissociation. During dimerisation, two benzoic-acid molecules combine to form one dimer, so the number of particles decreases and i becomes less than 1. Glucose has i approximately 1, while ionisation of NaCl or K₂SO₄ makes i greater than 1.

Boiling-point elevation is ΔT_b = iK_bm. Glucose is a non-electrolyte and has i = 1, whereas completely dissociated NaCl gives Na⁺ and Cl⁻, so i = 2. Because the two solutions have equal molality and the same solvent, solution B has twice the ideal particle effect and therefore the higher boiling point.

The van’t Hoff factor represents the ratio of the actual number of solute particles in solution to the number expected from the original formula. If two solute molecules associate to form one dimer, the number of particles becomes half, so i = 1/2 = 0.5. Therefore, complete dimer formation is the most suitable explanation. Dissociation would produce i greater than 1.

For a nonelectrolyte such as glucose, the van’t Hoff factor is i = 1. First calculate the moles: moles = 18/180 = 0.1 mol. Since the solvent mass is 1 kg, the molality is 0.1 mol kg⁻¹. Using ΔTf = iKf m, ΔTf = 1 × 1.86 × 0.1 = 0.186 K. Thus option A is correct.

Freezing-point depression is given by ΔTf = iKf m. When two solute molecules associate completely to form one dimer, the number of particles becomes half and the van’t Hoff factor becomes i = 0.5. At the same solvent and molality, the new depression is 0.5 × 0.372 = 0.186 K. Association therefore decreases the colligative effect.

Osmotic pressure is given by π = iCRT. Because concentration and temperature are identical for all solutions, the solution with the largest van’t Hoff factor will have the greatest osmotic pressure. Glucose gives 1 particle, NaCl gives 2, MgCl₂ gives 3, and Al₂(SO₄)₃ gives 5 ions on complete dissociation. Therefore, Al₂(SO₄)₃ has the highest osmotic pressure.

For a solution containing a non-volatile solute, Raoult’s law states that the vapour pressure of the solution is p = X_solvent × p°_solvent. The solute mole fraction is 0.05, so the solvent mole fraction is 1 − 0.05 = 0.95. Therefore, p = 0.95 × 100 mmHg = 95 mmHg. The vapour pressure decreases because the solute reduces the fraction of solvent molecules escaping into the vapour phase. Hence, option C is correct.

Boiling-point elevation is proportional to i m for a given solvent: ΔTb = iKb m. For the nonelectrolyte, i = 1 and the effective molality is 0.2 m. Completely dissociated NaCl has i = 2, so its effective molality is 2 × 0.1 = 0.2 m. Both solutions therefore have the same elevation, 0.102 K.

The depression in freezing point is ΔTf = iKf m. At equal molality and in the same solvent, the greatest depression corresponds to the largest van’t Hoff factor and therefore the lowest freezing point. Urea gives 1 particle, NaCl gives 2, CaCl₂ gives 3, and K₄[Fe(CN)₆] gives 8 ions on complete dissociation, not 5. Hence option D is correct.

The relation between the normal molar mass M and the observed molar mass Mobserved is i = M/Mobserved. If the observed molar mass is half the normal value, Mobserved = M/2. Therefore, i = M/(M/2) = 2. A value greater than one indicates that the solute has dissociated into more particles in solution, so option C is correct.

For dilute solutions, osmotic pressure depends on the total concentration of dissolved particles, represented by iC. Glucose does not dissociate, so iC = 1 × 0.2 = 0.2 M. NaCl completely dissociates into two ions, so iC = 2 × 0.1 = 0.2 M. Both sides therefore have equal osmotic pressure, making them isotonic. Hence, there is no net flow of water and option C is correct.

Adding more non-volatile solute particles lowers the mole fraction of the solvent and therefore lowers the solvent’s vapour pressure. In contrast, the relative lowering of vapour pressure, elevation in boiling point, and depression in freezing point increase with the number of solute particles. The question asks for the property itself that decreases, not the amount of its lowering. Thus, option D is correct.

Assume one mole of solute molecules initially. Of these, 75% remain as individual molecules, contributing 0.75 particles. The remaining 25% molecules associate into trimers; therefore, they form 0.25/3 = 0.0833 trimer particles. The total effective number of particles is 0.75 + 0.0833 = 0.8333. Hence, the van’t Hoff factor is approximately 0.833, so option C is correct. Association lowers i below 1.

For a non-electrolyte, the depression in freezing point is ΔTf = Kf m, where m is the molality, defined as moles of solute per kilogram of solvent. Since both solutions contain the same amount of solute, the solution with less solvent has the greater molality. Solution A has only 0.5 kg solvent, while B has 1.0 kg, so A has twice the molality and the greater freezing-point depression. Option A is correct.

Osmotic pressure is preferred for determining the molar mass of proteins and other macromolecules because it can be measured in very dilute solutions. The osmotic-pressure relation π = CRT gives a measurable quantity proportional to concentration, while boiling-point elevation, freezing-point depression, and relative vapour-pressure lowering may be extremely small for a dilute solution containing a high-molar-mass solute. Also, osmotic-pressure measurements do not require heating the protein.

Osmotic pressure is the preferred colligative property for finding the molar mass of proteins and other macromolecules in very dilute solutions. It follows π = cRT or π = CRT, so even a small concentration can produce a measurable osmotic pressure. The changes in boiling point, freezing point, or vapour pressure are often too small for accurate measurement when the solute has a very high molar mass. Osmotic measurements also avoid heating and possible protein decomposition. Hence, option C is correct.

Raoult’s law states that, for a solution containing a non-volatile solute, the lowering in the solvent vapour pressure is related to the solute mole fraction. Specifically, (p° − p)/p° = Xsolute for a dilute solution. Here p° is the vapour pressure of pure solvent and p is that of the solution. Therefore, option A is correct.

A van’t Hoff factor below 1 indicates association, because association decreases the number of independently moving solute particles. Benzoic acid associates to form dimers in the non-polar solvent benzene, so its effective particle number is reduced and i becomes less than 1. Glucose is a nonelectrolyte with i approximately 1, while NaCl and CaCl₂ generally dissociate and have i greater than 1.

The depression in freezing point is given by ΔTf = iKf m. Assuming the solute, solvent, and temperature remain comparable, i and Kf are unchanged, so ΔTf is proportional to molality. The mass of solute remains 5 g, but the solvent mass doubles from 100 g to 200 g; therefore molality becomes half. Consequently, the freezing-point depression also becomes half. Option B is correct.

For a solution showing association or dissociation, the freezing-point depression is calculated using ΔTf = iKf m. Substituting the given values gives ΔTf = 2.5 × 1.86 × 0.2 K. First, 1.86 × 0.2 = 0.372; multiplying by 2.5 gives 0.930 K. Thus the depression in freezing point is 0.930 K, so option C is correct.

The depression in freezing point is ΔTf = iKf m. At the same molality and with the same solvent, equal depression requires equal van’t Hoff factor i. Completely dissociated NaCl produces two ions, Na+ and Cl−, while KNO₃ produces K+ and NO₃−, so both have i approximately equal to 2. The other pairs produce different numbers of particles.

For boiling-point elevation, ΔTb = iKb m. If solute molecules associate, the number of independent particles decreases and i becomes less than 1. The actual boiling-point elevation is therefore smaller than expected for the original molecular formula. If i is ignored during calculation, the solute appears to have fewer moles, so its calculated molar mass is higher than the true value. Association is correct.