6: Molar Mass Determination

For a solute that dissociates, the relation between true and observed molar mass is M_obs = M_true/i. Therefore, M_true = i × M_obs = 1.4 × 75 = 105 g mol⁻¹. AB produces two ions on complete dissociation, so its van’t Hoff factor is i = 1 + α, where α is the fractional degree of dissociation. Hence α = 1.4 − 1 = 0.4, or 40%.

In colligative-property measurements, the observed molar mass is M observed = M true/i because the van’t Hoff factor changes the effective number of particles. Substituting the given values gives M observed = 90/0.6 = 150 g mol⁻¹. Since i is less than one, association has reduced particle number and makes the observed molar mass larger than the true value. Therefore D is correct.

Freezing-point depression is ΔTf = iKf m. With i = 1 and Kf for water = 1.86 K kg mol⁻¹, the molality is m = 0.0186/1.86 = 0.010 mol kg⁻¹. The solvent mass is 500 g = 0.5 kg, so solute moles are 0.010 × 0.5 = 0.005 mol. Thus M = 1.0/0.005 = 200 g mol⁻¹, so C is correct.

For dissociation, the observed molar mass is related to the true molar mass by M_obs = M_true/i. Thus M_obs = 132/2.2 = 60 g mol⁻¹. One AB₂ formula unit gives three particles on complete dissociation, so i = 1 + 2α. Substituting i = 2.2 gives 2.2 = 1 + 2α, hence α = 0.6, or 60% dissociation.

Colligative properties depend on the total number of dissolved particles, not merely on the formula units initially added. Dissociation increases the number of particles and gives i greater than 1, whereas association decreases the number and gives i less than 1. Consequently, an uncorrected calculation may produce an abnormal observed molar mass. The van’t Hoff factor is therefore required to relate observed and true molar masses correctly, making option B the only valid statement.

The moles of water are 18/18 = 1.00 mol. If ns is the number of moles of solute, its mole fraction is ns/(1 + ns) = 0.05. Solving gives ns = 0.05/0.95 = 0.05263 mol. The molar mass is therefore mass/moles = 2.0/0.05263 = 38.0 g mol⁻¹ approximately. Hence, option A is the correct answer.

In the freezing-point method, ΔTf = Kf m and m = (mass of solute × 1000)/(molar mass × mass of solvent in grams). Rearranging shows that calculated molar mass is directly proportional to the solvent mass used. The student uses 0.125 g instead of 125 g, which is 1000 times smaller. Therefore, the calculated molar mass will also be 1000 times smaller, so option B is correct.

For freezing-point depression, ΔTf = iKf m. Thus, the effective molality is 0.2325/1.86 = 0.125 mol kg⁻¹. Because i = 1.25, the actual molality of the solute is 0.125/1.25 = 0.100 mol kg⁻¹. The solvent mass is 500 g = 0.5 kg, so moles of solute are 0.100 × 0.5 = 0.050 mol. Therefore, molar mass = 2.5/0.050 = 50 g mol⁻¹, so option B is correct.

The elevation in boiling point is given by ΔTb = iKb m. Therefore, the molality is m = 0.156/(0.75 × 0.52) = 0.400 mol kg⁻¹. The solvent mass is 400 g = 0.400 kg, so the amount of solute is 0.400 × 0.400 = 0.160 mol. Hence, the molar mass is mass/moles = 4.8/0.160 = 30 g mol⁻¹. Thus, option A is the only correct answer.

For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute: x₂ = n₂/(n₁ + n₂). Water contains n₁ = 72/18 = 4 mol. Thus, 0.04 = n₂/(4 + n₂), giving 0.16 + 0.04n₂ = n₂ and n₂ = 0.1667 mol. The solute molar mass is 3/0.1667 ≈ 18 g mol⁻¹, so option A is correct.

For AB dissociation into two ions, i = 1 + α = 1 + 0.25 = 1.25. Using ΔTf = iKf m, the actual molality is m = 0.558/(1.25 × 1.86) = 0.240 mol kg⁻¹. The water mass is 300 g = 0.300 kg, so the solute amount is 0.240 × 0.300 = 0.072 mol. Therefore, the true molar mass is 3.6/0.072 = 50 g mol⁻¹, making option B correct.

Osmotic pressure follows π = iCRT. Hence, the true molar concentration is C = π/(iRT) = 0.615/(1.25 × 0.0821 × 300) ≈ 0.0200 mol L⁻¹. The solution volume is 500 mL = 0.500 L, so moles of solute are 0.0200 × 0.500 = 0.0100 mol. Consequently, the true molar mass is 2/0.0100 = 200 g mol⁻¹. Therefore, option C is correct.

A colligative property responds to the effective concentration of particles, which is represented by i × m. Here, i × m = 0.6 × 0.3 = 0.18 mol kg⁻¹. If the correction for i is ignored, this effective molality is mistaken for the actual molality. Therefore, the apparent molality is 0.18 m, making option A correct.

Use the colligative-property equation ΔT_f = iK_fm, where m is molality. Moles of solute = 4/100 = 0.04 mol, and the solvent mass is 200 g = 0.200 kg. Therefore, m = 0.04/0.200 = 0.20 mol kg⁻¹. Hence, ΔT_f = 1.5 × 1.86 × 0.20 = 0.558 K. Thus, option D is correct. The actual freezing point decreases by this amount.

For elevation of boiling point, use ΔT_b = iK_bm. The amount of solute is 3/60 = 0.05 mol, while 250 g of solvent equals 0.250 kg. Thus, the molality is m = 0.05/0.250 = 0.20 mol kg⁻¹. Substitution gives ΔT_b = 2 × 0.52 × 0.20 = 0.208 K. Therefore, option B is correct. The solution boils 0.208 K above the pure solvent.

The van’t Hoff factor compares the actual number of solute particles with the number expected from the formula. When i < 1, the solution has fewer independent particles than expected. This commonly results from association, such as dimerization or trimerization, where several solute molecules combine into fewer particles. Hence option B is correct.

For a non-electrolyte, boiling-point elevation is ΔTb = Kb m, where molality m equals moles of solute divided by kilograms of solvent. The solute mass and therefore its moles remain constant, but halving the solvent mass halves the denominator. Thus molality doubles, and the boiling-point elevation also doubles. Option A is correct.

For osmotic pressure, π = iCRT. Dissociation increases the number of solute particles, so i becomes greater than one and the measured osmotic pressure is larger than expected for the same analytical concentration. If the ordinary formula i = 1 is used, this unusually large pressure gives a calculated molar mass smaller than the true value. Therefore, option B is correct.

Freezing-point depression follows ΔTf = iKf m. Association combines solute molecules and reduces the number of independent particles, so i becomes less than one and the observed depression is smaller than the ideal value. If this reduced depression is interpreted without correcting for association, the formula gives fewer apparent moles and therefore a molar mass greater than the true value. Option B is correct.

For comparable non-electrolytes, ΔTf = Kf m, and molality is proportional to the number of solute moles per fixed solvent mass. With equal solute masses, the substance having the lower molar mass supplies more moles and hence more particles. It produces the larger freezing-point depression. Therefore, the first solute has the lower molar mass, so option B is correct.

Biomolecules such as proteins and polymers generally have very high molar masses, so only a small number of molecules is present even in a measurable sample. They may also decompose when heated, making boiling-point or freezing-point methods unsuitable. Osmotic pressure can be measured at room temperature with very dilute solutions, making it safer and more sensitive for biomolecules. Option A is correct.

Use ΔTf = Kf m and m = (wsolute/M)/(wsolvent in kg). Therefore, M = Kf × wsolute × 1000/(ΔTf × wsolvent in g). Substitution gives M = (1.86 × 2 × 1000)/(0.186 × 100) = 200 g mol⁻¹. The factor 1000 converts the solvent mass from grams to kilograms. Hence option B is correct.

For boiling-point elevation, ΔTb = Kb m, where m = (wsolute/M)/(wsolvent in kg). Rearranging gives M = Kb × wsolute × 1000/(ΔTb × wsolvent in g). Thus M = (2.6 × 1.5 × 1000)/(0.78 × 50) = 100 g mol⁻¹. Therefore, the molar mass is 100 g mol⁻¹ and option A is correct.

Relative lowering of vapour pressure is derived from Raoult’s law for a dilute solution containing a non-volatile solute. Because the solute does not contribute appreciably to the vapour phase, the lowering depends on its mole fraction. Under this assumption, the measured lowering can be related to the number of solute moles and hence to its molar mass. A volatile solute or a concentrated solution would require additional corrections.

In the freezing-point depression method, a dilute solution may lower the solvent’s freezing point by only a small fraction of a degree. Accurate molar-mass calculation therefore requires careful temperature measurement, good thermal contact, and avoidance of supercooling. The other listed procedures are not standard colligative-property methods for determining molar mass and do not depend on measuring such a tiny temperature difference.