6: Molar Mass Determination

Boiling-point elevation is expressed as ΔT_b = K_b m, where K_b is the ebullioscopic constant and m is the molality of the solution. For w₂ grams of solute with molar mass M, the amount is w₂/M moles. For w₁ grams of solvent, the solvent mass in kilograms is w₁/1000. Substitution and rearrangement give M = K_b w₂ × 1000/(ΔT_b w₁). Therefore, option A is correct. The other expressions either invert the relationship or omit the required mass and unit-conversion factors.

For a dilute solution, osmotic pressure obeys the van’t Hoff equation πV = nRT. If w grams of solute are present and its molar mass is M, then the number of moles is n = w/M. Substituting this into the gas-like equation gives πV = (w/M)RT, which can be rearranged to calculate M as M = wRT/(πV). Therefore, option A is the correct relation. Pressure, volume, temperature, and the gas constant must be expressed in compatible units.

For a dilute solution, depression in freezing point is ΔTf = Kf × (w₂ × 1000)/(M × w₁). Rearranging gives M = (Kf × w₂ × 1000)/(ΔTf × w₁). Substitution gives M = (1.86 × 1 × 1000)/(0.186 × 100) = 1860/18.6 = 100 g mol⁻¹. The factor 1000 converts the solvent mass from grams to kilograms, so option A is correct.

For boiling-point elevation, ΔTb = Kb × (w₂ × 1000)/(M × w₁). Therefore, M = (Kb × w₂ × 1000)/(ΔTb × w₁). Substituting the data gives M = (0.52 × 2 × 1000)/(0.104 × 100) = 1040/10.4 = 100 g mol⁻¹. The 1000 factor accounts for converting grams of solvent to kilograms. Hence, option B is correct.

Colligative properties depend on the number of dissolved particles rather than on the chemical identity, colour, or smell of the solute. For freezing-point depression, ΔT_f = iK_fm; for a non-dissociated solute, i = 1, so ΔT_f = K_fm. The measured depression gives the molality, and molality combined with the known solute mass and solvent mass gives the number of moles and hence molar mass. Thus particle number is the fundamental basis.

For a non-dissociated solute, the van’t Hoff factor is one, so ΔT_f = K_fm. Therefore, m = 0.279/1.86 = 0.15 mol kg⁻¹. The solvent mass is 200 g = 0.200 kg, so moles of solute = 0.15 × 0.200 = 0.030 mol. The molar mass is mass divided by moles: M = 3/0.030 = 100 g mol⁻¹. Hence option C is correct. Converting the solvent mass to kilograms is essential because molality uses kg of solvent.

For a non-dissociated solute, boiling-point elevation follows ΔT_b = K_bm. Hence the molality is m = 0.13/0.52 = 0.25 mol kg⁻¹. The solvent mass is 100 g = 0.100 kg, so the amount of solute is 0.25 × 0.100 = 0.025 mol. Therefore, molar mass M = 2.5 g/0.025 mol = 100 g mol⁻¹, making option C correct. The calculation must use solvent mass, not total solution mass, in the molality expression.

For a dilute solution of a non-dissociated solute, osmotic pressure is π = CRT, where C is molarity. Thus C = π/(RT) = 0.738/(0.082 × 300) = 0.030 mol L⁻¹. The solution volume is 250 mL = 0.250 L, so moles of solute = 0.030 × 0.250 = 0.0075 mol. Therefore, M = 1.8 g/0.0075 mol = 240 g mol⁻¹. Option C is correct. The volume must be expressed in litres because R is given in L atm units.

Use the osmotic-pressure equation π = CRT for a non-dissociated solute. The molarity is C = 0.984/(0.082 × 300) = 0.040 mol L⁻¹. Since 500 mL = 0.500 L, the number of moles in the solution is n = 0.040 × 0.500 = 0.020 mol. Hence the molar mass is M = mass/n = 4 g/0.020 mol = 200 g mol⁻¹. Therefore option C is correct. No van’t Hoff correction is needed because the solute does not dissociate.

Molality is defined as moles of solute per kilogram of solvent: m = n₂/m₁. Rearranging gives n₂ = m × m₁ = 0.02 mol kg⁻¹ × 0.5 kg = 0.010 mol. The molar mass is the mass of solute divided by its amount in moles, so M = 1.2 g/0.010 mol = 120 g mol⁻¹. Thus option C is correct. The solvent mass is already in kilograms, so no further conversion is required.

The molality relation is m = moles of solute divided by kilograms of solvent. Therefore, the moles of solute are n = 0.12 mol kg⁻¹ × 0.5 kg = 0.06 mol. Using M = mass/n, the molar mass is M = 6 g/0.06 mol = 100 g mol⁻¹. Hence option C is correct. The statement that the solute is non-dissociated confirms that the stated molality corresponds directly to the actual solute molecules, with no van’t Hoff-factor correction needed.

Assuming the solute is non-dissociated, ΔT_f = K_fm. Thus the molality is m = 0.186/1.86 = 0.10 mol kg⁻¹. The water mass is 50 g = 0.050 kg, so the moles of solute are n = 0.10 × 0.050 = 0.0050 mol. Therefore, M = 0.5 g/0.0050 mol = 100 g mol⁻¹, so option C is correct. The kilogram conversion for water is essential because the unit of K_f contains kg mol⁻¹.

For a non-dissociated solute, the boiling-point elevation is given by ΔT_b = K_b m, where m is molality. Thus, m = 0.104/0.52 = 0.20 mol kg^-1. The solvent mass is 150 g = 0.150 kg, so moles of solute = 0.20 × 0.150 = 0.030 mol. Therefore, molar mass = 1.5/0.030 = 50 g mol^-1. Hence, option B is correct.

For a non-dissociated solute, ΔT_b = K_b m. Therefore, the molality is m = 0.104/0.52 = 0.20 mol kg^-1. The solvent mass is 200 g, or 0.200 kg. Hence, moles of solute = 0.20 × 0.200 = 0.040 mol. The molar mass is mass divided by moles: M = 4/0.040 = 100 g mol^-1. Thus, option B is the only correct answer.

Osmotic pressure is the most suitable colligative property for determining the molar mass of polymers because it can be measured accurately in very dilute solutions. For a dilute solution, π = cRT or π = wRT/MV, so the molar mass can be calculated from a measurable pressure. In contrast, the changes in boiling point, freezing point, and vapour pressure are usually extremely small for dilute polymer solutions and are more difficult to measure reliably. Therefore, option D is correct.

For a non-electrolyte in dilute solution, osmotic pressure is given by π = CRT. Hence C = π/(RT) = 0.492/(0.082 × 300) = 0.020 mol L⁻¹. Because the solution volume is 1 L, the amount of solute is 0.020 mol. Therefore, molar mass M = mass/moles = 2 g/0.020 mol = 100 g mol⁻¹. Thus, option B is correct. The same result follows directly from π = wRT/(MV).

Because the solute does not dissociate, its van’t Hoff factor is i = 1, so π = CRT. The concentration is C = 0.738/(0.082 × 300) = 0.030 mol L⁻¹. The volume is 500 mL = 0.500 L, so moles of solute = 0.030 × 0.500 = 0.015 mol. Therefore, M = 3.6/0.015 = 240 g mol⁻¹. Hence, option C is correct. Converting millilitres to litres is essential.

First convert the volume: 250 mL = 0.250 L. The number of moles is obtained from n = Molarity × volume, so n = 0.05 mol L⁻¹ × 0.250 L = 0.0125 mol. The molar mass is therefore mass divided by amount: M = 2 g/0.0125 mol = 160 g mol⁻¹. Thus, option C is correct. Using 250 instead of 0.250 in the molarity calculation would produce an erroneous result because molarity uses litres.

Large molecules such as polymers and biomolecules form very dilute solutions when used for molar-mass measurements. In such solutions, the changes in boiling point and freezing point are extremely small, and vapour-pressure lowering is also difficult to measure accurately. Osmotic pressure, however, can be measured as a pressure even in a very dilute solution and is related to molar mass by π = wRT/MV. Therefore, the osmotic pressure method does not require measurement of a tiny temperature change, making option A correct.

For a dilute solution of a non-electrolyte, freezing-point depression is related to molality by ΔTf = Kf m, so molality is obtained first from the measured temperature change. Next, moles of solute are calculated by multiplying molality by the solvent mass in kilograms. Finally, the molar mass is found by dividing the given solute mass by its moles. Therefore, option A gives the correct logical sequence.

Osmotic pressure is related to molarity by π = CRT. Therefore, the first step is to calculate the solute molarity C from the measured osmotic pressure, temperature, and gas constant. The moles are then obtained from n = CV, with V expressed in litres. Finally, molar mass is calculated using M = w/n. Molality, Kf, and Kb belong to other concentration or colligative methods, so option A is correct.

For a non-dissociated solute, ΔTf = Kf m. Hence, the molality is m = 0.093/1.86 = 0.05 mol kg⁻¹. The solvent mass is 100 g = 0.1 kg, so the moles of solute are n = 0.05 × 0.1 = 0.005 mol. Therefore, M = 1 g/0.005 mol = 200 g mol⁻¹, making option C correct. No van’t Hoff correction is needed because the solute is non-dissociated.

Use the boiling-point relation ΔTb = Kb m to find molality: m = 0.052/0.52 = 0.1 mol kg⁻¹. Convert the solvent mass: 250 g = 0.25 kg. The amount of solute is therefore n = 0.1 × 0.25 = 0.025 mol. Its molar mass is M = 2 g/0.025 mol = 80 g mol⁻¹. Hence, option C is correct; the solvent mass must be used in kilograms.

For elevation of boiling point, ΔT_b = K_b m, where m is molality. Therefore, m = 0.208/0.52 = 0.40 mol kg⁻¹. Molality equals moles of solute divided by kilograms of solvent, so moles of solute = 0.40 × 0.25 = 0.10 mol. Hence option B is correct. The solvent mass, not the total solution mass, is used in this calculation.

Molality is defined as moles of solute per kilogram of solvent. Thus, moles of solute = molality × solvent mass = 0.4 mol kg⁻¹ × 0.25 kg = 0.10 mol. Molar mass is mass divided by amount in moles, so M = 8 g/0.10 mol = 80 g mol⁻¹. Since the solute is non-dissociated, no van’t Hoff factor correction is required. Option C is correct.