6: Molar Mass Determination

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.

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.

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.

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 \(M_{\text{calc}}=M_{\text{actual}}/1.20=5M_{\text{actual}}/6\).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.