6: Molar Mass Determination

First convert the volume into litres: 250 mL = 0.250 L. Molarity is moles per litre of solution, so moles of solute = M × V = 0.02 mol L⁻¹ × 0.250 L = 0.005 mol. The molar mass is mass divided by moles: 1.0 g/0.005 mol = 200 g mol⁻¹. Therefore, option C is the only correct answer.

Molality is moles of solute per kilogram of solvent. Therefore, the moles of solute are n = 0.1 mol kg⁻¹ × 0.3 kg = 0.03 mol. The molar mass is then calculated as mass divided by moles: M = 3.6 g/0.03 mol = 120 g mol⁻¹. Because the solute is non-dissociated, the ordinary molality relation applies without any particle-number correction. Option C is correct.

Convert the solvent mass into kilograms: 200 g = 0.200 kg. Using molality = moles of solute/kg of solvent, the moles of solute are 0.1 mol kg⁻¹ × 0.200 kg = 0.020 mol. The molar mass is therefore 2 g/0.020 mol = 100 g mol⁻¹. Since molality uses solvent mass, the 200 g water value must be converted before calculation. Option C is correct.

First convert the solvent mass: 300 g = 0.300 kg. The molality relation gives moles of solute = 0.05 mol kg⁻¹ × 0.300 kg = 0.015 mol. The molar mass is the solute mass divided by its amount: 1.5 g/0.015 mol = 100 g mol⁻¹. Therefore, the calculated molar mass corresponds uniquely to option C.

Molar mass is calculated by dividing the mass of the solute by the number of moles: M = mass / moles. Therefore, M = 2.4 g / 0.012 mol = 200 g mol⁻¹. The osmotic-pressure method has already provided the number of moles, so no additional osmotic-pressure formula is needed. Hence, option C is correct.

Molality is defined as moles of solute per kilogram of solvent. In freezing-point depression and boiling-point elevation calculations, using solvent mass in grams instead of kilograms introduces a factor-of-1000 error. This greatly changes the calculated number of moles and therefore produces a very wrong molar mass. The other choices are not calculation errors.

For osmotic-pressure determination of molar mass, M = wRT/(πV), where w, R, T, and V are constant in this question. Thus, molar mass M is inversely proportional to osmotic pressure π. If π changes from π to 2π, the new molar mass becomes wRT/(2πV), which is half of the original value. Therefore, option B is correct.

For a dilute solution, osmotic pressure is given by π = nRT/V. Since n = w/M, the molar-mass formula is M = wRT/(πV). Substituting the data gives M = (2 × 0.082 × 300)/(0.6 × 1) = 82 g mol⁻¹. The units are consistent because pressure is in atm and R is in L atm K⁻¹ mol⁻¹. Thus, option A is correct.

For a non-electrolyte, ΔTb = Kb wB/MB, so rearrangement gives MB = Kb wB/ΔTb. The temperature difference is in the denominator. If ΔTb is measured too small while all other quantities are unchanged, division by the smaller denominator gives a larger calculated molar mass. Therefore, the result is greater than the actual value.

For freezing-point depression, ΔTf = Kf m, and molality m is (wB/MB)/(wA/1000). Rearranging gives MB = Kf wB × 1000/(ΔTf wA). Substitution gives MB = (2.0 × 1.0 × 1000)/(0.20 × 100) = 100 g mol⁻¹. The 1000 factor converts the solvent mass from grams to kilograms. Therefore, option B is correct.

For a dilute solution of a non-electrolyte, the elevation in boiling point is ΔT_b = K_b m, where m is molality. Using the mass form, M = (K_b × w_solute × 1000)/(ΔT_b × w_solvent). Therefore, M = (0.5 × 2 × 1000)/(0.25 × 100) = 40 g mol⁻¹. Hence option B is correct. The solvent mass must be expressed in grams in this formula, while K_b already contains kg-based units.

Molality is defined using the mass of solvent only: m = moles of solute divided by kilograms of solvent. The total solution mass includes both solute and solvent, so substituting it in the denominator changes the concentration used in ΔTf = iKf m. Consequently, the calculated molality and any molar mass obtained from it will be erroneous. Therefore, A is correct.

The standard equations used for molar-mass determination from colligative properties assume a dilute, nearly ideal solution. In a concentrated solution, solute particles interact more strongly with one another and with solvent molecules, and the simple proportional relationships may no longer hold accurately. Activity effects and changes in solvent properties can therefore produce a systematic error in the calculated molar mass. Using a sufficiently dilute solution improves reliability.

For a non-dissociating solute, freezing-point depression is ΔT_f = K_f m, where m is molality. Keeping the solute mass unchanged keeps the number of solute moles unchanged. Doubling the solvent mass doubles the denominator in molality, so the molality becomes half its original value. Since K_f remains constant for the same solvent, ΔT_f also becomes half. Therefore, option B is correct.

For a dilute solution of a non-electrolyte, ΔT_b = K_b m. Using the mass form of molality, the molar mass M is M = K_b × w_solute × 1000 ÷ (ΔT_b × w_solvent). Substitution gives M = (0.5 × 3 × 1000) ÷ (0.30 × 250) = 1500 ÷ 75 = 20 g mol⁻¹. Hence option B is correct.

In colligative-property equations, the symbols \(w_A\) and \(w_B\) represent specified masses, usually the solvent and solute respectively, according to the convention used in the formula. Interchanging them changes the numerator, denominator, or mass ratio. The resulting value will therefore not represent the intended molar mass. The symbols must be assigned exactly as defined before substitution. Hence, option A is correct.

For a non-dissociating solute, osmotic pressure is given by \(\pi=CRT\), where \(C\) is molar concentration, \(R\) is the gas constant, and \(T\) is absolute temperature. At the same temperature, \(R\) and \(T\) are common to both solutions, so osmotic pressure is directly proportional to \(C\). A greater osmotic pressure therefore means a higher molar concentration. Hence, option A is correct.

The experimentally calculated molar mass is half the actual molar mass, which means that the observed colligative effect is greater than the effect expected from undissociated molecules. Dissociation breaks solute particles into two or more particles, increasing the number of particles in solution. Since colligative properties depend on particle number, the enhanced effect gives a lower apparent molar mass. Therefore, dissociation is the possible process.

The experimental molar mass is greater than the actual molar mass, indicating that the solution shows a smaller colligative effect than expected. Association joins two or more solute molecules to form larger particles, reducing the total number of independent particles. Because colligative properties depend on the number of solute particles, this reduction makes the calculated or apparent molar mass higher. Hence, association of the solute is the suitable conclusion.

Boiling-point elevation and freezing-point depression are both colligative properties. For dilute solutions, their relations are ΔTb = Kb m and ΔTf = Kf m, where m is the molality of the solution. Molality uses the moles of solute per kilogram of solvent, so the mass of solvent is important in both methods. Neither method requires a membrane, and the solute is generally considered non-volatile.

For a completely non-dissociating and non-associating solute, the van’t Hoff factor is i = 1 because each formula unit remains one particle in solution. Using i = 2 incorrectly assumes that every solute unit produces two effective particles. This overestimates the particle concentration and changes the colligative-property equation, so the molar mass calculated from the observation becomes erroneous. Therefore, option A is correct.

For a dilute solution, osmotic pressure is given by π = wRT/(MV), so M = wRT/(πV). Convert the volume first: 500 mL = 0.5 L. Substitution gives M = (1.5 × 0.082 × 300)/(0.738 × 0.5) = 49? Actually, the numerator is 36.9 and the denominator is 0.369, giving M = 100 g mol⁻¹. Therefore, option C is correct. Careful unit conversion is essential.

In colligative-property methods, such as boiling-point elevation and freezing-point depression, molar mass is calculated using expressions like M = Kb w₂ × 1000/(ΔTb w₁) or M = Kf w₂ × 1000/(ΔTf w₁). The temperature difference is in the denominator. Consequently, even a small error in ΔT changes the calculated value appreciably, especially when ΔT is small. Accurate temperature measurement is therefore essential.

For a nonelectrolyte, ΔTf = Kf m, and m = (w₂/M)/(w₁ in kg). Hence M = Kf w₂ × 1000/(ΔTf w₁), where w₂ is solute mass and w₁ is solvent mass in grams. Substitution gives M = (1.86 × 1.5 × 1000)/(0.279 × 100) = 2790/27.9 = 100 g mol⁻¹. Therefore, option B is correct.

For a nonelectrolyte, the boiling-point relation is ΔTb = Kb m. With solute mass w₂ and solvent mass w₁ in grams, the molar mass is M = Kb w₂ × 1000/(ΔTb w₁). Substituting the data gives M = (0.52 × 2 × 1000)/(0.052 × 200) = 1040/10.4 = 100 g mol⁻¹. The temperature value used is the elevation, not the total boiling temperature.