6: Molar Mass Determination

Correct answer: C. 80 g mol⁻¹

Explanation: For a non-dissociated solute, i = 1 and ΔTf = Kf m. Therefore, the molality is m = 0.465/1.86 = 0.25 mol kg−1. The solvent mass is 200 g = 0.200 kg, so moles of solute = 0.25 × 0.200 = 0.050 mol. Its molar mass is mass/moles = 4/0.050 = 80 g mol−1. Hence, option C is correct.

Correct answer: C. 200 g mol⁻¹

Explanation: For a dilute solution, osmotic pressure is π = nRT/V, and n = w/M. Therefore, M = wRT/(πV). Convert 250 mL to 0.250 L, then substitute: M = (2.0 × 0.082 × 300)/(0.984 × 0.250) = 200 g mol⁻¹. Correct volume conversion is essential because R uses litres. Hence, option C is correct.

Correct answer: C. \(120\,\mathrm{g\,mol^{-1}}\)

Explanation: The van't Hoff factor relates normal and observed molar masses by \(i=M_{\text{normal}}/M_{\text{observed}}\). Therefore, \(M_{\text{normal}}=iM_{\text{observed}}=1.5\times80=120\,\mathrm{g\,mol^{-1}}\). Thus, option C is correct. Because \(i>1\), the solute produces more particles than expected, so its observed molar mass is lower than its normal molar mass.

Correct answer: A. \(\frac{5}{6}\) of the actual molar mass

Explanation: For boiling-point elevation, \(\Delta T_b=K_bm\), and for fixed solute and solvent masses, molality is proportional to \(1/M\). Hence, the calculated molar mass is inversely proportional to the measured \(\Delta T_b\). A 20% overestimate makes the measured value \(1.20\) times actual, so \(M_{\text{calc}}=M_{\text{actual}}/1.20=5M_{\text{actual}}/6\).

Correct answer: A. It will be lower

Explanation: For freezing-point depression, \(M_B=K_fw_B\times1000/(\Delta T_fw_A)\), where \(w_A\) is the solvent mass. The incorrect value changes the denominator from 100 to 120 g, making it 1.2 times larger. Therefore, the calculated molar mass becomes \(100/120=5/6\) of the actual value, so it is lower. Option A is correct.

Correct answer: A. Both have equal molar masses

Explanation: For a non-dissociating solute, the depression in freezing point is ΔTf = Kf m. Because the solvent is the same, Kf is identical for both solutions; equal ΔTf therefore means equal molality. The solvent masses are equal, so the numbers of solute moles are equal. Since the solute masses are also equal, M = mass/moles gives equal molar masses. The conclusion depends on equal solvent masses and absence of association or dissociation.

Correct answer: B. 0.67

Explanation: For association into trimers, three monomer particles combine to form one particle. If α is the fraction associated, the van’t Hoff factor is i = 1 − 2α/3 because each trimerisation event decreases the particle count by two. With α = 0.50, i = 1 − (2 × 0.50)/3 = 1 − 1/3 = 2/3 ≈ 0.67. The factor is below one because association reduces the number of particles in solution.

Correct answer: C. 120 g mol⁻¹

Explanation: The observed freezing-point depression gives i × m = ΔTf/Kf = 0.186/1.86 = 0.10 mol kg⁻¹. For AB dissociating by 50%, i = 1 + α = 1.5, so the actual molality of AB is 0.10/1.5 = 0.0667 mol kg⁻¹. With 0.500 kg water, moles of solute = 0.0667 × 0.500 = 0.0333 mol. Therefore, M = 4/0.0333 ≈ 120 g mol⁻¹. Option C is correct.

Correct answer: D. 300 g mol⁻¹

Explanation: Use the osmotic-pressure equation π = iCRT, where C is the molarity of the original solute. Thus C = 0.246/(0.5 × 0.082 × 300) = 0.020 mol L⁻¹. The volume is 0.200 L, so moles of solute formula units = 0.020 × 0.200 = 0.004 mol. Therefore, the true molar mass is 1.2/0.004 = 300 g mol⁻¹. Option D is correct.

Correct answer: B. 100 g mol⁻¹

Explanation: First calculate the effective molality: i m = ΔTf/Kf = 0.279/1.86 = 0.15 mol kg⁻¹. Since i = 1.5, the true solute molality is m = 0.15/1.5 = 0.10 mol kg⁻¹. The solvent mass is 0.300 kg, so moles of solute = 0.10 × 0.300 = 0.030 mol. Hence, the true molar mass is 3/0.030 = 100 g mol⁻¹. Therefore, option B is correct.

Correct answer: D. \(400\,g\,mol^{-1}\)

Explanation: Use the osmotic-pressure equation \(\pi=iCRT\). Thus, \(C=\pi/(iRT)=1.23/(2\times0.082\times300)=0.025\,mol\,L^{-1}\). The volume is \(0.250\,L\), so the amount of solute is \(0.025\times0.250=0.00625\,mol\). Therefore, the true molar mass is \(2.5/0.00625=400\,g\,mol^{-1}\). Hence, option D is correct.

Correct answer: C. \(80\,g\,mol^{-1}\)

Explanation: Freezing-point depression follows \(\Delta T_f=iK_fm\), where \(m\) is the true molality. The effective molality is \(0.372/1.86=0.2\,mol\,kg^{-1}\). Since \(i=2\), the true molality is \(0.2/2=0.1\,mol\,kg^{-1}\). For \(0.200\,kg\) of water, moles of solute are \(0.1\times0.200=0.020\). Hence, molar mass \(=1.6/0.020=80\,g\,mol^{-1}\), so C is correct.

Correct answer: B. \(75\,g\,mol^{-1}\)

Explanation: One molecule of \(AB_2\) dissociates into three particles, so for degree of dissociation \(\alpha\), the van’t Hoff factor is \(i=1+\alpha(3-1)=1+2\alpha\). With \(\alpha=0.50\), \(i=1+2(0.50)=2\). The observed molar mass is \(M_{\text{true}}/i=150/2=75\,g\,mol^{-1}\). Therefore, option B is correct.

Correct answer: C. \(112.5\,g\,mol^{-1}\)

Explanation: For dimerisation, two monomer particles form one dimer, so the van’t Hoff factor is \(i=1-\alpha/2\). With \(\alpha=0.40\), \(i=1-0.40/2=0.80\). Association decreases the number of particles, so the observed molar mass becomes larger: \(M_{\text{observed}}=M_{\text{true}}/i=90/0.80=112.5\,g\,mol^{-1}\). Thus, option C is correct.

Correct answer: A. \(50\,g\,mol^{-1}\)

Explanation: Boiling-point elevation is given by \(\Delta T_b=iK_bm\). The effective molality is \(0.052/0.52=0.1\,mol\,kg^{-1}\). Because \(i=0.5\), the true molality is \(0.1/0.5=0.2\,mol\,kg^{-1}\). The solvent mass is \(0.500\,kg\), so moles of solute are \(0.2\times0.500=0.1\). Therefore, molar mass \(=5/0.1=50\,g\,mol^{-1}\), making A correct.

Correct answer: B. 100 g mol⁻¹

Explanation: For a nonelectrolyte, ΔTf = Kf m and the van’t Hoff factor is 1. Using the molar-mass form, M = Kf × w₂ × 1000 ÷ (ΔTf × w₁), where w₂ is solute mass in grams and w₁ is solvent mass in grams. Thus M = 1.86 × 2.4 × 1000 ÷ (0.1488 × 300) = 100 g mol⁻¹. Option B is correct.

Correct answer: A. 48 g mol⁻¹

Explanation: For a nonelectrolyte, boiling-point elevation follows ΔTb = Kb m. In terms of molar mass, M = Kb × w₂ × 1000 ÷ (ΔTb × w₁), with w₂ = 3.6 g solute and w₁ = 250 g solvent. Therefore M = 0.52 × 3.6 × 1000 ÷ (0.156 × 250) = 48 g mol⁻¹. Hence option A is correct.

Correct answer: C. 300 g mol⁻¹

Explanation: Use the dilute-solution osmotic-pressure equation π = nRT/V = wRT/(MV). Rearranging gives M = wRT/(πV). Convert 500 mL to 0.500 L, then substitute: M = (0.75 × 0.082 × 300)/(0.123 × 0.500) = 300 g mol⁻¹. The volume must be the volume of the solution, so option C is correct.

Correct answer: C. 120 g mol−1

Explanation: The depression in freezing point is ΔTf = Tf° − Tf = 273.000 − 272.814 = 0.186 K. For a nonelectrolyte, ΔTf = Kf m, and using the gram form, M = Kf × wsolute × 1000/(ΔTf × wsolvent). Therefore, M = (1.86 × 1.2 × 1000)/(0.186 × 100) = 120 g mol−1. The absolute freezing point is not substituted for ΔTf; only the decrease is used.

Correct answer: B. 100 g mol−1

Explanation: First calculate the elevation in boiling point: ΔTb = Tb − Tb° = 373.104 − 373.000 = 0.104 K. For a nonelectrolyte, ΔTb = Kb m. Using M = Kb × wsolute × 1000/(ΔTb × wsolvent), we obtain M = (0.52 × 2.0 × 1000)/(0.104 × 100) = 100 g mol−1. The final boiling point itself is not used in place of the elevation.

Correct answer: B. 60 g mol−1

Explanation: For a nonelectrolyte, ΔTf = Kf m. In terms of masses, the molar mass is M = Kf × wsolute × 1000/(ΔTf × wsolvent). Substituting the data gives M = (1.86 × 1.8 × 1000)/(0.372 × 150). Since 1.86/0.372 = 5, this becomes 5 × 1.8 × 1000/150 = 60 g mol−1. The solvent mass must be expressed consistently in grams in this formula.

Correct answer: B. 40 g mol−1

Explanation: For a nonelectrolyte, ΔTb = Kb m. Using the mass form of the molality equation, M = Kb × wsolute × 1000/(ΔTb × wsolvent). Thus M = (0.52 × 2.5 × 1000)/(0.26 × 125). Because 0.52/0.26 = 2, the result is (2 × 2.5 × 1000)/125 = 40 g mol−1. The 125 g solvent mass is used directly with the factor 1000.

Correct answer: B. 600 g mol−1

Explanation: For a dilute polymer solution, osmotic pressure is π = cRT = (w/MV)RT, so M = wRT/(πV). Convert 250 mL to 0.250 L. Substitution gives M = (0.25 × 0.082 × 300)/(0.041 × 0.250) = 6.15/0.01025 = 600 g mol−1. Osmotic pressure is especially useful for polymers because it can determine their large molar masses without requiring vaporisation.

Correct answer: C. 100 g mol⁻¹

Explanation: For molar-mass determination from boiling-point elevation, use M = Kb w₂ × 1000/(ΔTb w₁), where w₂ is the solute mass in grams and w₁ is the solvent mass in grams. Substitution gives M = (0.52 × 2.0 × 1000)/(0.026 × 400) = 1040/10.4 = 100 g mol⁻¹. The small temperature change is correctly handled through the denominator, so option C is the answer.

Correct answer: C. 200 g mol⁻¹

Explanation: For a dilute nonelectrolyte solution, osmotic pressure is given by π = cRT = (w/MV)RT. Rearranging gives M = wRT/(πV). Substituting w = 0.6 g, R = 0.082 L atm K⁻¹ mol⁻¹, T = 300 K, π = 0.369 atm, and V = 0.2 L gives M = (0.6 × 0.082 × 300)/(0.369 × 0.2) = 200 g mol⁻¹. Therefore, option C is correct.