1.4 g of acetone dissolved in 100 g of benzene gave a solution that freezes at 277.12 K. Pure benzene freezes at 278.4 K. 2.8 g of a solid (A) dissolved in 100 g of benzene gave a solution that froze at 277.76 K. Calculate the molecular mass of (A).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 232
A non-volatile solute lowers the freezing point of a solvent. The depression of freezing point is proportional to the molality of the solute, which gives the molar mass of the solute:
\( \displaystyle \Delta T_f=K_f\,m \qquad M_2=\frac{K_f\times w_2\times 1000}{\Delta T_f\times w_1} \)
Kf of benzene is not given, so it is first found from the acetone solution, whose molar mass is known.
Acetone, \( \ce{CH3COCH3} \) (\( \ce{C3H6O} \)): M = 3(12) + 6(1) + 16 = 58 g mol−1.
\( \displaystyle \Delta T_f=278.4-277.12=1.28\,\mathrm{K} \)
\( \displaystyle K_f=\frac{\Delta T_f\times M_2\times w_1}{w_2\times 1000}=\frac{1.28\times 58\times 100}{1.4\times 1000} \)
\( \displaystyle K_f=5.303\,\mathrm{K\,kg\,mol^{-1}} \)
Now the solution of (A) in the same solvent:
\( \displaystyle \Delta T_f=278.4-277.76=0.64\,\mathrm{K} \)
\( \displaystyle M_A=\frac{5.303\times 2.8\times 1000}{0.64\times 100} \)
\( \displaystyle M_A=232\,\mathrm{g\,mol^{-1}} \)
The molecular mass of (A) is 232.
Shortcut: Both solutions use 100 g of benzene, so Kf cancels and ΔTf is proportional to w2/M2:
\( \displaystyle M_A=58\times\frac{2.8}{1.4}\times\frac{1.28}{0.64}=58\times 2\times 2=232 \)