6: Molar Mass Determination

Correct answer: A. Decrease in freezing point

Explanation: The freezing-point-depression method is based on the colligative property ΔTf. The freezing point of the pure solvent and that of the solution are measured, and their difference, ΔTf = Tf° − Tf, is calculated. This depression is related to molality by ΔTf = Kf m. Using the known mass of solute and solvent, the molar mass can then be determined. The other listed observations do not enter this method.

Correct answer: A. Solvent

Explanation: Kf is called the molal depression constant or cryoscopic constant. For a given solvent, it has a characteristic value and represents the depression in freezing point produced by a one-molal solution of a non-electrolyte. Therefore, Kf depends on the nature of the solvent, not on the identity of the solute, thermometer, or container. Changing water to another solvent changes the value of Kf.

Correct answer: A. 1

Explanation: A non-dissociating solute remains as the same molecular species after dissolving. Consequently, one solute molecule gives one effective particle, so the ratio of actual particles to expected particles is one. The van’t Hoff factor is therefore i = 1. This is the normal ideal value for a non-electrolyte that neither dissociates into ions nor associates to form larger molecules.

Correct answer: A. Less than normal

Explanation: Dissociation converts each solute unit into two or more ions, increasing the number of particles in solution. Since colligative properties depend on the number of particles, the measured effect becomes larger than expected for an undissociated solute. If this larger effect is interpreted without correcting for dissociation, the calculated number of moles is too high and the observed molar mass, mass divided by calculated moles, becomes lower than the normal value.

Correct answer: A. More than normal

Explanation: Association occurs when two or more solute particles combine to form a single larger species. The total number of effective particles therefore decreases, so the observed colligative effect is smaller than the effect expected for the same amount of unassociated solute. If this smaller effect is used without correcting for association, the calculated number of moles is too low. Since molar mass equals mass divided by moles, the observed molar mass becomes greater than the normal value.

Correct answer: A. Because they depend on the number of solute particles

Explanation: Colligative properties, such as osmotic pressure, boiling-point elevation, and freezing-point depression, depend mainly on the number of dissolved solute particles rather than their chemical identity. From the measured change, the amount of solute in moles can be calculated. Comparing this amount with the known mass gives molar mass. Therefore, option A is correct.

Correct answer: A. Osmotic pressure method

Explanation: A semipermeable membrane allows solvent molecules to pass through while restricting solute particles. This selective movement is the basis of osmosis. In the osmotic-pressure method, the pressure required to stop the net movement of solvent through such a membrane is measured. The other listed colligative-property methods measure temperature or vapour-pressure changes and do not require a membrane. Hence, option A is correct.

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

Explanation: ΔT_b represents the elevation in boiling point caused by dissolving a non-volatile solute. It is calculated as ΔT_b = T_b(solution) − T_b°(pure solvent). Because the solution usually boils at a higher temperature than the pure solvent, this difference is positive. It is a temperature difference, not the mass, volume, or colour of the solution. Therefore, option A is correct.

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

Explanation: ΔT_f denotes freezing-point depression. It is defined as ΔT_f = T_f°(pure solvent) − T_f(solution), where T_f° is the freezing point of the pure solvent and T_f is the freezing point of the solution. A dissolved solute usually lowers the freezing point, so the difference is positive. Option D refers to boiling-point change, not freezing-point depression. Thus, option A is correct.

Correct answer: D. Osmotic pressure

Explanation: Osmotic pressure is most suitable for determining the molar mass of proteins and other macromolecules because it can be measured in very dilute solutions. The method does not require heating or freezing, which helps prevent denaturation of sensitive biomolecules. Using π = CRT, the concentration of the macromolecule can be related to its molar mass. The other changes may be too small in dilute solutions. Therefore, option D is correct.

Correct answer: C. 60 g mol⁻¹

Explanation: Molar mass is the mass of one mole of a substance and is calculated using M = m/n. Here, the mass m is 30 g and the amount n is 0.5 mol. Therefore, M = 30 g ÷ 0.5 mol = 60 g mol⁻¹. Option C is correct. The value 30 g mol⁻¹ would incorrectly treat the given mass as the mass of one mole rather than 0.5 mol.

Correct answer: A. Kilogram

Explanation: Molality is defined as the number of moles of solute present per kilogram of solvent: m = moles of solute / mass of solvent in kg. Consequently, the solvent mass must be expressed in kilograms before calculation. If the mass is given in grams, it must be divided by 1000. Volume in litres belongs to molarity, not molality. Hence, option A is correct.

Correct answer: A. Mass of the solvent, 500 g

Explanation: In the usual notation for solution formulas, w_A denotes the mass of the solvent, while w_B commonly denotes the mass of the solute. In this problem, the solvent has a mass of 500 g and the solute has a mass of 10 g. Therefore, w_A = 500 g. The total solution mass is 510 g, but it is not represented by w_A. Thus, option A is correct.

Correct answer: A. Molal elevation constant

Explanation: K_b is the ebullioscopic constant, also called the molal elevation constant. It appears in the relation ΔT_b = K_b m for a dilute solution, where ΔT_b is the elevation in boiling point and m is the molality. The symbol K_f is associated with freezing-point depression and is called the molal depression constant. Therefore, option A is correct.

Correct answer: A. Molal depression constant

Explanation: K_f is called the cryoscopic constant or molal depression constant. It is used in the equation ΔT_f = K_f m for a dilute solution, where ΔT_f is the depression in freezing point and m is the molality of the solute. By contrast, K_b is the molal elevation constant for boiling-point elevation. Therefore, option A is correct.

Correct answer: A. Relative lowering of vapour pressure

Explanation: Relative lowering of vapour pressure is a colligative property because it depends on the number of solute particles rather than their chemical identity. For a dilute solution, the relative lowering is related to the mole fraction of the solute. From this relation, the number of solute moles and then its molar mass can be calculated. Mass percentage, density, and viscosity are not colligative properties.

Correct answer: A. It will increase

Explanation: For a dilute solution, the osmotic-pressure relation gives M = wRT/(πV), where w is the solute mass, R is the gas constant, T is absolute temperature, π is osmotic pressure, and V is solution volume. If R, T, π, and V remain constant, M is directly proportional to w. Therefore, increasing the solute mass increases the calculated molar mass.

Correct answer: A. It will decrease

Explanation: The osmotic-pressure equation for molar mass is M = wRT/(πV). When the solute mass, gas constant, temperature, and solution volume are fixed, the molar mass is inversely proportional to osmotic pressure. Thus, increasing π increases the denominator and reduces the value of M. It cannot become zero merely because osmotic pressure increases.

Correct answer: A. It decreases

Explanation: For a non-electrolyte solute, the boiling-point method can be written as M_B = K_b × 1000 × w_B/(ΔT_b × w_A), where K_b is the ebullioscopic constant, w_B is solute mass, and w_A is solvent mass. With all other quantities fixed, ΔT_b is in the denominator. Therefore, a larger boiling-point elevation gives a smaller calculated molar mass.

Correct answer: A. It will be higher

Explanation: For freezing-point depression, the molar mass is calculated from M_B = K_f × 1000 × w_B/(ΔT_f × w_A). For fixed solvent and solute masses, M_B is inversely proportional to ΔT_f. If the measured depression is smaller, the denominator used in the calculation is smaller, so the calculated molar mass becomes artificially higher than it should be. This is a measurement-effect question.

Correct answer: A. Mass of solute, 1 g

Explanation: In the standard notation used for colligative-property formulas, component B commonly denotes the solute and component A denotes the solvent. Therefore, wB means the mass of the solute, which is 1 g in this example. The 100 g quantity is wA, the solvent mass, while 101 g would be the total solution mass and is not represented by wB.

Correct answer: A. Kelvin scale

Explanation: The osmotic-pressure equation is π = nRT/V, or π = wRT/(MV). Because the gas constant R is used in an absolute-temperature relationship, T must be expressed in kelvin. Celsius or Fahrenheit values cannot be substituted directly. For example, 27 °C must first be converted to 300 K, using K = °C + 273 approximately.

Correct answer: A. 300 K

Explanation: Temperature in the osmotic-pressure equation must be absolute temperature in kelvin. To convert from Celsius, add 273 to the Celsius value: T = 27 + 273 = 300 K. Using 27 K would incorrectly treat the Celsius number as kelvin, while 273 K corresponds approximately to 0 °C. Therefore, the correct answer is 300 K.

Correct answer: A. So that the vapour-pressure change is due mainly to the solvent

Explanation: A non-volatile solute has negligible vapour pressure and does not significantly enter the vapour phase. Consequently, the vapour above the solution is produced essentially by the solvent, and the measured lowering of vapour pressure can be related to the solute mole fraction. This relationship is required for determining the solute molar mass accurately. A volatile solute would interfere with the measurement.

Correct answer: A. 0.2 K

Explanation: The elevation in boiling point is defined as ΔT_b = T_b(solution) − T_b(pure solvent). Substitution gives ΔT_b = 100.2 °C − 100.0 °C = 0.2 °C. A temperature difference has the same numerical value on the Celsius and Kelvin scales, so ΔT_b = 0.2 K. Therefore, option A is correct; the absolute boiling point should not be confused with the temperature difference.