6: Molar Mass Determination

Correct answer: C. \(200\,g\,mol^{-1}\)

Explanation: For a nonelectrolyte, \(\Delta T_f=K_fm\), so the molality is \(m=0.093/1.86=0.05\,mol\,kg^{-1}\). Since the solvent mass is exactly \(1\,kg\), the amount of solute is \(0.05\,mol\). The molar mass is therefore mass divided by moles: \(M=10/0.05=200\,g\,mol^{-1}\). The assumption of no association or dissociation means \(i=1\). Hence, option C is correct.

Correct answer: C. 200 g mol⁻¹

Explanation: Convert the volume first: 400 mL = 0.400 L. The amount of solute is n = molarity × volume = 0.015 mol L⁻¹ × 0.400 L = 0.006 mol. Molar mass is mass divided by amount: M = 1.2 g ÷ 0.006 mol = 200 g mol⁻¹. Therefore, option C is correct. The osmotic-pressure method is useful because osmotic pressure can provide the solution concentration, from which molar mass is calculated when the solute mass and solution volume are known.

Correct answer: A. A measurable osmotic pressure can be obtained even in very dilute solutions.

Explanation: Large molecules have high molar masses, so a given mass contains very few moles. Consequently, their boiling-point elevation or freezing-point depression may be too small for accurate measurement. Osmotic pressure is measurable even at very low concentrations and is proportional to molar concentration, making it especially useful for biomolecules and polymers. Thus, option A is correct.

Correct answer: A. 0.1488 K

Explanation: For the same solute, solvent, and temperature conditions, ΔTf is proportional to the ratio of solute mass to solvent mass. The ratio of the new concentration effect to the old one is (2.0/250)/(1.0/100) = 0.8. Therefore, the new depression is 0.186 × 0.8 = 0.1488 K. Both solute and solvent masses must be included in the comparison; comparing solute masses alone would be incorrect.

Correct answer: B. 36 g mol⁻¹

Explanation: For a dilute solution containing a non-volatile solute, the relative lowering of vapour pressure is approximately equal to the mole fraction of the solute, and therefore Δp/p⁰ ≈ n₂/n₁. The moles of water are n₁ = 9.0/18 = 0.50 mol. Hence n₂ = 0.10 × 0.50 = 0.050 mol. The solute molar mass is M₂ = mass/moles = 1.8/0.050 = 36 g mol⁻¹. Thus option B is correct.

Correct answer: C. 100 g mol⁻¹

Explanation: Molality is defined as the number of moles of solute present in one kilogram of solvent. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.06 × 0.25 = 0.015 mol. The molar mass is then the mass of solute divided by its amount: M = 1.5 g/0.015 mol = 100 g mol⁻¹. The solvent mass, not solution mass, must be used in molality. Hence option C is correct.

Correct answer: B. 1200 g mol⁻¹

Explanation: For a dilute polymer solution, osmotic pressure is given by π = (w/MV)RT, where w is the polymer mass, M its molar mass, and V the solution volume in litres. Rearranging gives M = wRT/(πV). Substituting w = 0.2 g, T = 300 K, V = 0.200 L, π = 0.0205 atm, and R = 0.082 gives M = (0.2 × 0.082 × 300)/(0.0205 × 0.200) ≈ 1200 g mol⁻¹. Therefore, option B is correct.

Correct answer: B. 100 g mol⁻¹

Explanation: For a nonelectrolyte, the freezing-point relation is ΔT_f = K_f m, where m is the molality of the solute. Using the molar-mass form, M = K_f × w₂ × 1000/(ΔT_f × w₁), with w₂ = 1.0 g solute and w₁ = 50 g camphor. Therefore M = 40 × 1.0 × 1000/(8 × 50) = 100 g mol⁻¹. This is the principle used in the Rast method. Hence option B is correct.

Correct answer: B. 100 g mol⁻¹

Explanation: For a non-electrolyte, the depression in freezing point is given by ΔTf = Kf m, where m is the molality of the solution. Using the mass form, M = (Kf × w₂ × 1000)/(ΔTf × w₁), where w₂ is the mass of solute in grams and w₁ is the mass of solvent in grams. Substituting the data: M = (1.86 × 1.5 × 1000)/(0.1116 × 250) = 2790/27.9 = 100 g mol⁻¹. Therefore, option B is correct. The calculation assumes no association or dissociation of solute particles.

Correct answer: D. 120 g mol⁻¹

Explanation: The van’t Hoff factor connects normal and observed molar masses through i = Mnormal/Mobserved. Hence, Mnormal = i × Mobserved. Using i = 2.4 and Mobserved = 50 g mol⁻¹, Mnormal = 2.4 × 50 = 120 g mol⁻¹. The factor greater than one indicates dissociation or an increase in the number of solute particles, so option D is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: Molality is the number of moles of solute per kilogram of solvent. The solvent mass is 500 g = 0.500 kg. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.05 × 0.500 = 0.025 mol. The molar mass is mass/moles = 3.0/0.025 = 120 g mol⁻¹, so option C is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: For a dilute solution of a non-electrolyte, osmotic pressure is given by π = (w/MV)RT, where w is the mass of solute, M is its molar mass, V is the solution volume in litres, and T is the absolute temperature. Rearranging gives M = wRT/(πV). Substituting the data, M = (2.2 × 0.082 × 300)/(0.451 × 1.0) = 54.12/0.451 ≈ 120 g mol⁻¹. Therefore, option C is correct.

Correct answer: C. 80 g mol⁻¹

Explanation: For a dilute nonelectrolyte solution, ΔTf = Kf m, where m is molality. Using m = (w₂/M)/(w₁ in kg), the molar mass formula is M = Kf w₂ × 1000/(ΔTf w₁), with w₁ in grams. Substitution gives M = (2.0 × 4.0 × 1000)/(0.4 × 250) = 80 g mol⁻¹. Hence option C is correct.

Correct answer: C. 0.16 m

Explanation: For the same solute, molality is proportional to the mass of solute and inversely proportional to the mass of solvent: m = (mass of solute/M) ÷ mass of solvent. The solute mass changes from 2.0 g to 4.0 g, so it doubles, while the solvent changes from 0.5 kg to 0.25 kg, so it is halved. Therefore the molality becomes 2 × 2 = 4 times the original value. New molality = 0.04 × 4 = 0.16 m.

Correct answer: A. Molality will be extremely low and the calculated molar mass may be much too high

Explanation: The conversion 100 g = 0.100 kg is essential in the molality formula, m = moles of solute per kilogram of solvent. If 100 kg is used instead of 0.100 kg, the solvent mass is taken as 1000 times too large, so the calculated molality becomes 1000 times too small. Any molar mass obtained from this erroneous molality will consequently be unreliable; in the usual rearrangement, the molar mass may appear excessively high.

Correct answer: B. 1000 times smaller

Explanation: The correct solvent mass is 250 g = 0.250 kg. Treating it as 250 kg makes the solvent mass 250/0.250 = 1000 times too large. In the freezing-point method, ΔTf = Kf m and the number of solute moles is calculated from m × kilograms of solvent. Thus the erroneous calculation gives 1000 times too many moles. Since molar mass equals mass of solute divided by its moles, the calculated molar mass becomes 1000 times smaller than the correct value.

Correct answer: B. 80 g mol⁻¹

Explanation: For freezing-point depression, ΔT_f = iK_fm, where m is the actual molality of solute particles. Therefore, m = 0.279/(1.5 × 1.86) = 0.100 mol kg⁻¹. The solvent mass is 500 g = 0.5 kg, so moles of solute = 0.100 × 0.5 = 0.050 mol. Hence, true molar mass = 4/0.050 = 80 g mol⁻¹. Thus option B is correct.

Correct answer: A. 50 g mol⁻¹

Explanation: The boiling-point relation is ΔT_b = iK_bm. Thus the actual molality is m = 0.156/(0.75 × 0.52) = 0.400 mol kg⁻¹. The solvent mass is 300 g = 0.300 kg, so the amount of solute is 0.400 × 0.300 = 0.120 mol. Therefore, the true molar mass is 6/0.120 = 50 g mol⁻¹, which is option A.

Correct answer: A. 19.6 g mol⁻¹

Explanation: For a non-volatile solute, relative lowering of vapour pressure equals the solute mole fraction: x₂ = n₂/(n₁+n₂) = 0.02. Water moles are n₁ = 90/18 = 5 mol. Thus n₂ = 0.02(5+n₂), giving n₂ = 0.1/0.98 = 0.102 mol approximately. The molar mass is 2/0.102 ≈ 19.6 g mol⁻¹, so option A is correct.

Correct answer: D. 240 g mol⁻¹

Explanation: Osmotic pressure follows π = iCRT, where C is the true molarity. Therefore, C = 1.845/(1.5 × 0.0821 × 300) ≈ 0.050 M. In 0.250 L, moles of solute are n = 0.050 × 0.250 = 0.0125 mol. Hence the true molar mass is 3/0.0125 = 240 g mol⁻¹. Therefore, option D is correct.

Correct answer: B. 75 g mol⁻¹

Explanation: The effective molality is related to the true molality by m_effective = i × m_true. Therefore, m_true = 0.05/0.75 = 0.0667 mol kg⁻¹. The solvent mass is 0.3 kg, so moles of solute = 0.0667 × 0.3 = 0.020 mol. Hence, molar mass = 1.5 g/0.020 mol = 75 g mol⁻¹. Thus, option B is correct. Since i is less than 1, the solute undergoes association.

Correct answer: B. i = 1.5, dissociation

Explanation: For abnormal molar masses, the relation is observed molar mass = true molar mass/i. Therefore, i = true molar mass/observed molar mass = 150/100 = 1.5. A van’t Hoff factor greater than 1 means that the number of solute particles increases in solution. This increase is normally caused by dissociation, such as an electrolyte separating into ions. Hence, option B is correct.

Correct answer: C. 120 g mol⁻¹

Explanation: Colligative properties give an observed molar mass related to the true molar mass by M observed = M true/i. Rearranging gives M true = i × M observed. Substitution gives M true = 2.4 × 50 = 120 g mol⁻¹. Because i is greater than one, the solute produces additional particles, consistent with dissociation. Therefore option C is correct.

Correct answer: A. i = 1.5, dissociation

Explanation: The osmotic-pressure equation is π = iCRT, so i = π/(CRT). Here CRT = 0.01 × 0.082 × 300 = 0.246 atm, and i = 0.369/0.246 = 1.5. Since i is greater than one, the number of solute particles has increased, indicating dissociation rather than association or normal non-electrolyte behaviour. Thus option A is correct.

Correct answer: C. 400 g mol⁻¹

Explanation: For osmotic pressure, π = iCRT. Since i = 1, C = π/(RT) = 0.123/(0.082 × 300) = 0.005 mol L⁻¹. The solution volume is 0.100 L, so solute moles equal 0.005 × 0.100 = 0.0005 mol. Hence molar mass = mass/moles = 0.2/0.0005 = 400 g mol⁻¹, making C correct.