6: Molar Mass Determination

For a dilute solution of a nonelectrolyte, osmotic pressure follows πV = nRT. Since n = w/M, rearrangement gives M = wRT/(πV). Substituting the values, M = (0.6 × 0.082 × 300)/(0.246 × 0.2) = 14.76/0.0492 = 300 g mol⁻¹. The litre unit of volume is consistent with the given value of R.

Molality is defined as moles of solute per kilogram of solvent. The mass of water is 200 g = 0.200 kg. Thus, moles of solute = 0.05 mol kg⁻¹ × 0.200 kg = 0.010 mol. The molar mass is mass divided by moles: M = 1.8 g/0.010 mol = 180 g mol⁻¹. Therefore, option B is correct.

For a nonelectrolyte, ΔTf = Kf m. Using masses in grams, the molar-mass formula is M = Kf w₂ × 1000/(ΔTf w₁). Here w₂ = 3 g and w₁ = 500 g, so M = (1.86 × 3 × 1000)/(0.186 × 500) = 5580/93 = 60 g mol⁻¹. The 500 g solvent mass must be used exactly as given, not replaced by 100 g.

For elevation of boiling point, ΔTb = Kb m, where m is the molality of the solute. For a solute of mass w₂ dissolved in w₁ grams of solvent, the molar mass is M = Kb × w₂ × 1000 ÷ (ΔTb × w₁). Substitution gives M = 0.52 × 1 × 1000 ÷ (0.052 × 100) = 100 g mol⁻¹. Therefore, option B is correct.

For a dilute solution, osmotic pressure is given by π = nRT/V, or π = CRT when concentration is used. With n, R, and T constant, osmotic pressure is inversely proportional to volume. Therefore, reducing the volume from V to V/2 makes π equal to nRT/(V/2) = 2nRT/V, so the osmotic pressure doubles. Hence option B is correct.

The osmotic-pressure equation for a solution showing possible association or dissociation is π = iCRT, where i is the van’t Hoff factor. Rearranging gives i = π/(CRT). Substituting the data, i = 2.46 ÷ (0.1 × 0.082 × 300) = 2.46 ÷ 2.46 = 1. A value of one indicates no effective change in the number of solute particles, so option B is correct.

For a nonelectrolyte, the freezing-point depression is ΔTf = Kf m. If w₂ is the solute mass and w₁ is the solvent mass in grams, M = Kf × w₂ × 1000 ÷ (ΔTf × w₁). Substituting the data gives M = 1.86 × 4.5 × 1000 ÷ (0.372 × 250) = 90 g mol⁻¹. Hence option B is correct.

The boiling-point elevation relation for a nonelectrolyte is ΔTb = Kb m. In terms of masses, M = Kb × w₂ × 1000 ÷ (ΔTb × w₁), where w₂ is solute mass and w₁ is solvent mass in grams. Therefore, M = 0.52 × 2.4 × 1000 ÷ (0.104 × 200) = 60 g mol⁻¹. Thus option A is correct.

For a dilute solution, osmotic pressure is π = nRT/V = wRT/(MV), where w is the solute mass, M is its molar mass, and V is the solution volume. Rearranging gives M = wRT/(πV). Substitution gives M = (0.9 × 0.082 × 300)/(0.246 × 0.3) = 22.14/0.0738 = 300 g mol⁻¹. Therefore, option C is correct. The volume must be expressed in litres because R is given in L atm units.

For a dilute polymer solution, osmotic pressure is related to molar mass by π = wRT/(MV), where w is the solute mass, M is its molar mass, V is the solution volume, and T is temperature. Convert 250 mL to 0.250 L. Therefore, M = wRT/(πV) = (0.5 × 0.082 × 300)/(0.082 × 0.250) = 600 g mol⁻¹. Hence option B is correct.

Molar mass is calculated by dividing the mass of the substance by its amount in moles: M = w/n. Here, w = 18 g and n = 0.2 mol. Therefore, M = 18/0.2 = 90 g mol⁻¹. Thus, option B is correct. The value 180 g mol⁻¹ would result from incorrectly using 0.1 mol instead of 0.2 mol. In solution chemistry, this calculation converts the measured number of solute moles into the mass of one mole.

Osmotic pressure is especially useful for finding the molar masses of proteins, polymers, and other macromolecules. Even a dilute solution of a large molecule can produce a measurable osmotic pressure, while heating is unnecessary; this helps prevent decomposition. Other colligative methods may produce changes that are too small or may require heating. Therefore, option A is the only suitable general answer.

Complete dissociation increases the number of particles in solution. Consequently, the van’t Hoff factor becomes greater than one (i > 1). For colligative-property calculations, the observed molar mass is related to the true value by M_observed = M_true/i, so it is smaller when i > 1. Association has i < 1 and gives an apparently larger molar mass. Hence, option B is correct.

Assume that 100 original molecules are present. Twenty-five molecules participate in dimerisation, so they produce 25/2 = 12.5 dimer particles. The remaining 75 molecules remain single. Thus, the final number of particles is 75 + 12.5 = 87.5, and i = 87.5/100 = 0.875. Therefore, option C is correct. Association decreases the particle number, so i must be less than one.

When solute particles associate, two or more particles combine to form fewer effective particles. The van’t Hoff factor then becomes less than one (i < 1). Since M_observed = M_true/i, division by a value below one makes the observed molar mass greater than the true molar mass. Dissociation produces the opposite effect, with i > 1. Hence, association in option A is correct.

Molality is defined as moles of solute per kilogram of solvent. Here, 100 g of solvent equals 0.100 kg. Therefore, moles of solute = 0.30 × 0.100 = 0.030 mol. The molar mass is mass divided by moles, so M = 3 g ÷ 0.030 mol = 100 g mol⁻¹. The essential step is converting the solvent mass from grams to kilograms before using the molality formula.

Osmotic pressure is most suitable for high-molar-mass solutes such as proteins because it can be measured in very dilute solutions at ordinary temperature. For a solute of large molar mass, the changes in boiling point, freezing point, or vapour pressure are often extremely small and difficult to measure accurately. The relation π = cRT, or M = wRT/(πV), is therefore useful for determining the molar mass.

A very dilute solution contains solute particles sufficiently far apart that solute–solute interactions become nearly negligible. Consequently, the solution behaves approximately ideally, and the relation π = cRT can be applied more accurately. Osmosis does occur, so option B is false; the temperature is not necessarily zero, and osmotic pressure must be measured. These ideal-dilute conditions make the calculated molar mass more reliable.

Polymers have extremely high molar masses, so a given mass of polymer contains very few molecules. Consequently, boiling-point elevation and freezing-point depression are often too small for precise measurement. Osmotic pressure, however, can be measured in dilute polymer solutions at ordinary temperature. Using π = cRT or πV = nRT allows the molar mass to be calculated. Osmotic pressure is not restricted to electrolytes; nonelectrolytes also produce it.

Osmotic pressure is preferred for proteins and other high-molar-mass solutes because it can be measured accurately in dilute solutions at ordinary temperature. For such substances, the number of molecules per unit mass is small, so their effects on vapour pressure, boiling point, and freezing point are usually very small. The osmotic-pressure relation π = cRT provides a practical way to calculate the molar mass without heating or freezing the sample.

The cryoscopic constant Kf and ebullioscopic constant Kb are characteristic properties of the solvent. They are not universal constants and may differ substantially from one solvent to another. Therefore, for freezing-point depression, the Kf value of the actual solvent must be used; for boiling-point elevation, the Kb value of that solvent must be used. Using the constant of a different solvent gives an incorrect molar mass.

Molality is defined as the moles of solute present per kilogram of solvent. First calculate the moles of solute: 2.5 g ÷ 50 g mol⁻¹ = 0.05 mol. Next convert the solvent mass: 500 g = 0.5 kg. Therefore, molality = 0.05 mol ÷ 0.5 kg = 0.10 mol kg⁻¹. Hence, option B is correct.

Molality is calculated using kilograms of solvent, not grams: m = moles of solute ÷ kilograms of solvent. If a mass in grams is used directly, the denominator becomes 1000 times too large numerically, producing a major error in the calculated molality and therefore in the molar mass. The other actions do not alter the calculation. Thus, option A is correct.

For a nonelectrolyte, the depression in freezing point is given by ΔT_f = K_f m, where m is the molality. Therefore, m = 0.60/2.0 = 0.30 mol kg⁻¹. The solvent mass must be converted to kilograms: 50 g = 0.050 kg. Moles of solute = 0.30 × 0.050 = 0.015 mol. Hence, molar mass = mass/moles = 1.5/0.015 = 100 g mol⁻¹. Thus, option B is correct.

For boiling-point elevation, ΔT_b = K_b m, where m is the molality of the solution. Thus, m = 0.30/0.75 = 0.40 mol kg⁻¹. The solvent mass is 75 g = 0.075 kg, so the amount of solute is 0.40 × 0.075 = 0.030 mol. The molar mass is therefore mass divided by moles: M = 3.0/0.030 = 100 g mol⁻¹. Hence, option B is the only correct answer. The solvent, not the solution, is used in the denominator of molality.