6: Molar Mass Determination

Correct answer: D. Osmotic pressure

Explanation: For polymers and other high-molar-mass solutes, osmotic pressure is the most useful colligative property because it can be measured in dilute solutions at ordinary temperature. The corresponding changes in boiling point, freezing point, or vapour pressure are often too small for precise measurement. Osmotic-pressure data can therefore provide a more dependable estimate of polymer molar mass.

Correct answer: C. 0.492 atm

Explanation: For a non-dissociating solute, osmotic pressure is given by π = CRT. At constant temperature, with the same solute and solvent conditions, π is directly proportional to concentration. The concentration changes from 0.01 M to 0.02 M, so it doubles. Therefore, the osmotic pressure also doubles: π2 = (0.02/0.01) × 0.246 = 0.492 atm. The value 0.246 atm would apply only to the original concentration.

Correct answer: C. \(90\,\mathrm{g\,mol^{-1}}\)

Explanation: Molar mass is the mass of a substance divided by the amount of substance in moles: \(M=m/n\). Here, \(m=9.0\,\mathrm{g}\) and \(n=0.10\,\mathrm{mol}\). Therefore, \(M=9.0/0.10=90\,\mathrm{g\,mol^{-1}}\). The reference to the colligative property simply supplies the amount of solute; it does not change the definition. Hence, option C is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: The amount of solute is already given directly as 0.04 mol, so the stated molality is not needed for finding molar mass. Apply M = mass/moles: M = 4.8 g ÷ 0.04 mol = 120 g mol⁻¹. Therefore, the correct choice is option C; no additional solvent-mass calculation is required.

Correct answer: B. 0.186 K

Explanation: For a non-dissociated solute, the van’t Hoff factor is i = 1, so the freezing-point depression is ΔT_f = iK_fm = K_fm. Moles of solute = 2.5/100 = 0.025 mol. The solvent mass is 250 g = 0.250 kg, giving molality m = 0.025/0.250 = 0.100 mol kg⁻¹. Therefore, ΔT_f = 1.86 × 0.100 = 0.186 K, so option B is correct.

Correct answer: B. 0.104 K

Explanation: For a non-dissociated solute, i = 1 and the boiling-point elevation is ΔT_b = iK_bm = K_bm. The amount of solute is 1.2/60 = 0.020 mol. The solvent mass is 100 g = 0.100 kg, so molality is m = 0.020/0.100 = 0.200 mol kg⁻¹. Thus, ΔT_b = 0.52 × 0.200 = 0.104 K. Therefore, option B is correct.

Correct answer: B. The volume of the entire solution

Explanation: The osmotic-pressure equation is obtained from πV = nRT, where V is the volume occupied by the solution containing the dissolved solute. Since osmotic pressure depends on the concentration of solute particles in the entire solution, V must be the total solution volume, not the solvent volume alone or the container’s empty space. Thus option B is correct.

Correct answer: A. The van’t Hoff factor is usually 1

Explanation: A non-electrolyte normally neither dissociates into ions nor associates significantly in the solution. Consequently, the number of solute particles remains close to the number predicted from its formula, so the van’t Hoff factor i is approximately 1. Colligative-property equations can then be used with the ordinary molar mass. The word “usually” is appropriate because experimental deviations may occur, but i is not automatically 2.

Correct answer: A. The difference between the boiling points of the solution and the pure solvent

Explanation: In this method, the boiling point of the pure solvent and the boiling point of the solution are measured experimentally. Their difference is the elevation in boiling point, written as ΔTb = Tb(solution) − Tb(pure solvent). For a dilute solution, ΔTb is related to molality by ΔTb = Kb m, so this measured temperature difference is then used to calculate the amount of solute and its molar mass. The other properties are irrelevant to the method.

Correct answer: A. A larger freezing-point depression is produced for the same molality

Explanation: For a dilute nonelectrolyte solution, the depression in freezing point is given by ΔTf = Kf m, or by ΔTf = iKf m when abnormal particle behaviour is included. If molality and the van’t Hoff factor remain unchanged, a larger cryoscopic constant Kf produces a larger temperature change. The depression is therefore easier to observe accurately, improving experimental sensitivity. It does not alter the solute mass or molar mass.

Correct answer: A. It remains unchanged

Explanation: Freezing-point depression is proportional to molality: ΔTf = Kf m. Molality is the number of moles of solute divided by the kilograms of solvent. Doubling the solute mass doubles its moles, while doubling the solvent mass doubles the denominator. Their ratio therefore remains unchanged, so the molality and hence the freezing-point depression remain unchanged, provided the solution stays dilute and no association or dissociation changes occur.

Correct answer: A. The moles of solute must be doubled

Explanation: For a dilute solution, osmotic pressure is described by πV = nRT, or π = nRT/V. If temperature T and volume V are constant, osmotic pressure is directly proportional to the number of moles n of solute. Consequently, doubling n doubles π. For an electrolyte or an associating solute, the effective relation includes the van’t Hoff factor, but with unchanged particle behaviour the required change is still to double the amount of solute.

Correct answer: A. It will become double

Explanation: For a dilute solution, the molar mass is calculated using M = K_b w_2 × 1000/(ΔT_b w_1), where the measured elevation, solute mass, and solvent mass are kept fixed. Since K_b appears in the numerator, replacing it by 2K_b multiplies the calculated M by 2. Thus the reported molar mass becomes double. This is a calculation error, not a real change in the solute’s molar mass; options B and D reverse or overstate the proportional effect.

Correct answer: A. It will be much lower

Explanation: The osmotic-pressure relation for molar mass is M = wRT/(πV), so temperature must be expressed on the absolute Kelvin scale and it is directly proportional to the calculated M. The correct temperature is 27 °C + 273 ≈ 300 K, whereas the student uses only 27 K. Therefore the calculated value is multiplied by 27/300, about 0.09, and is much lower than the correct value. It is not negative because all quantities in the formula remain positive.

Correct answer: A. Osmotic pressure

Explanation: Osmotic pressure is a colligative property because, for a dilute solution at fixed temperature, it depends on the concentration of dissolved particles: π = iCRT. Association or dissociation changes the effective particle concentration, so measuring π can reveal an abnormal molecular mass or allow its correction using i. Colour, smell, and taste are not colligative properties and cannot provide this basis.