6: Molar Mass Determination

Colligative properties depend on the number of dissolved solute particles rather than on their chemical identity. Measurements such as boiling-point elevation, freezing-point depression, or osmotic pressure allow us to calculate the amount of solute present. Combining the solute mass with its calculated number of moles gives molar mass through M = mass/moles.

Molar mass is defined as the mass of a substance divided by the amount of substance in moles. Therefore, M = m/n, where m is the sample mass and n is the number of moles. If m is measured in grams, M is expressed in g mol⁻¹. The other relations either reverse the ratio or use volume and therefore do not define molar mass.

Freezing-point depression is calculated using molality: ΔTf = Kf m, or ΔTf = Kf × (moles of solute)/(mass of solvent in kilograms). Since molality is defined as moles of solute per kilogram of solvent, the solvent mass must be converted to kilograms before substitution. Grams are not used directly in the molality formula.

The elevation in boiling point is represented by ΔTb = iKb m, where Kb is the molal elevation constant or ebullioscopic constant of the solvent. The symbol Kf belongs to freezing-point depression, R is the gas constant, and p° denotes the vapour pressure of the pure solvent. Thus, Kb is the only suitable constant.

The depression in freezing point is given by ΔTf = iKf m. Here, Kf is the molal freezing-point depression constant, also called the cryoscopic constant, and its value depends on the solvent. Kb is used for boiling-point elevation, R is the gas constant, and i is the van’t Hoff factor rather than a solvent constant.

Large biomolecules such as proteins and polymers generally have very high molar masses and are available only in dilute solutions. Osmotic pressure can be measured appreciably even in dilute solutions and at room temperature, while heating methods may decompose these substances. Therefore, the osmotic-pressure method is especially convenient and reliable for their molar-mass determination.

Molar mass is calculated from M = m/n, where m is the mass in grams and n is the amount in moles. Substituting the given values gives M = 4 g / 0.05 mol = 80 g mol⁻¹. Therefore, option C is correct. A value of 40 g mol⁻¹ would result from an incorrect division and would not satisfy the stated mass-to-mole ratio.

Molar mass is calculated by dividing the mass of a substance by the number of moles: M = m/n. Here, m = 2 g and n = 0.02 mol, so M = 2/0.02 = 100 g mol⁻¹. Therefore, option B is correct. The unit is grams per mole because mass is measured in grams and amount of substance in moles. This is a direct application of the mole concept and is useful in solution calculations.

Molar mass is defined as the mass of one mole of a substance. It is calculated using M = m/n, where m is the given mass and n is the amount in moles. Substituting the values gives M = 5 g / 0.05 mol = 100 g mol⁻¹. Therefore, option B is correct. Always retain the unit g mol⁻¹, since molar mass expresses grams of substance per mole.

One kilogram contains 1000 grams. Therefore, to convert 100 g into kilograms, divide by 1000: 100 g ÷ 1000 = 0.1 kg. Hence, option B is correct. This conversion is particularly important in molality calculations because molality is defined as moles of solute per kilogram of solvent. Using grams instead of kilograms would produce a value that is wrong by a factor of 1000.

Molar mass is calculated by dividing the mass of a substance by the number of moles: M = mass ÷ moles. Here, M = 8 g ÷ 0.10 mol = 80 g mol⁻¹. Therefore, option C is correct. The unit is g mol⁻¹ because grams are divided by moles. A common mistake is to multiply 8 by 0.10 instead of dividing.

A non-volatile solute has negligible vapour pressure and does not appreciably enter the vapour phase. Consequently, the vapour above the solution is produced essentially by the solvent, and the lowering of vapour pressure can be related to the number of dissolved solute particles. This relationship is used to calculate molar mass. Therefore, option A is correct.

The osmotic-pressure equation is π = CRT, where T represents absolute temperature. Absolute temperature must be expressed in kelvins, so any Celsius value must first be converted using T(K) = t(°C) + 273.15. Using Celsius directly would make the numerical calculation inconsistent with the gas constant and give an incorrect osmotic pressure or molar mass. Therefore, option B is correct.

To convert Celsius to kelvin, add approximately 273 to the Celsius temperature. Thus, T = 27 + 273 = 300 K. Using the more precise conversion, 27 + 273.15 = 300.15 K, which is approximately 300 K. Therefore, option C is correct. Kelvin temperature is used in osmotic-pressure and other thermodynamic equations because it is an absolute temperature scale.

For a non-dissociated solute, the van’t Hoff factor is i = 1, so the osmotic-pressure equation becomes π = CRT. Rearranging gives C = π ÷ RT. Substitution gives C = 0.246 ÷ (0.082 × 300) = 0.246 ÷ 24.6 = 0.010 mol L⁻¹, or 0.010 M. Hence, option B is correct. The units also confirm the result is molarity.

The osmotic-pressure method uses π = CRT, where C is molarity. Molarity is defined as moles of solute per litre of solution, so the solution volume must be expressed in litres when calculating C or the amount of solute. Freezing-point depression and boiling-point elevation commonly use molality, which is based on kilograms of solvent rather than litres of solution. Therefore, option A is correct.

Molality is defined as the number of moles of solute present per kilogram of solvent: m = moles of solute / mass of solvent in kg. Therefore, the solvent mass must specifically be converted into kilograms. Molarity instead uses the volume of the solution in litres, while colour and the name of a substance are not concentration units. Molality is especially used in colligative-property calculations such as elevation of boiling point and depression of freezing point.

Molality is calculated by the relation m = moles of solute / mass of solvent in kilograms. Rearranging gives moles of solute = molality × mass of solvent. Here, the number of moles is 0.2 mol kg−1 × 1 kg = 0.20 mol. Thus, option C is correct. The unit kg cancels properly, leaving the answer in moles.

Molar mass is the mass of one mole of a substance and is calculated using M = mass / number of moles. Substituting the given values gives M = 20 g / 0.2 mol = 100 g mol−1. Therefore, option B is correct. The unit is grams per mole because a mass in grams is divided by an amount of substance in moles. This relation is fundamental in molar-mass determination.

For a solution containing a non-volatile solute, Raoult’s law gives the relation (p° − p)/p° = xsolute, where p° is the vapour pressure of the pure solvent and p is the vapour pressure of the solution. Thus, the relative lowering of vapour pressure equals the solute mole fraction in a dilute ideal solution. Measuring it allows the number of solute moles, and hence molar mass, to be determined.

Molar mass is defined as the mass of a substance divided by its amount in moles: M = m/n. Therefore, knowing the solute mass m and the number of solute moles n is sufficient for direct calculation. The colour, smell, container shape, preparation time, and day do not provide the required mass-to-mole relationship. In experimental work, colligative properties may help determine the unknown number of moles.

One formula unit of NaCl produces two separate particles on complete dissociation: one Na+ ion and one Cl− ion. The van’t Hoff factor is the ratio of the actual number of solute particles in solution to the number expected without dissociation. Therefore, ideal complete dissociation gives i = 2. In real solutions, ion interactions may cause a value slightly different from 2, but the ideal value asked here is exactly 2.

When one formula unit of calcium chloride, CaCl₂, completely dissociates in water, it produces one Ca²⁺ ion and two Cl⁻ ions. Thus, the total number of particles formed is 1 + 2 = 3. For complete dissociation, the van’t Hoff factor equals the number of particles produced, so i = 3. This assumes ideal behaviour and no ion pairing.

Molality is defined as the number of moles of solute present in one kilogram of solvent. In freezing-point depression and boiling-point elevation calculations, the solvent mass must therefore be converted from grams to kilograms before using the formula. Forgetting this conversion introduces a factor of 1000 and produces a seriously incorrect molar mass. Writing units and reading the solute mass are not mistakes by themselves.

Large molecules, such as proteins and other biomolecules, are usually studied in very dilute solutions. Their boiling-point elevation and freezing-point depression may then be extremely small and difficult to measure accurately. Osmotic pressure can be measured effectively even for such dilute solutions and is directly related to concentration through π = cRT. Therefore, osmotic pressure is the most convenient colligative property for determining the molar mass of large molecules.