When 0.44 g of a substance is dissolved in 22.2 g of benzene, the freezing point of benzene is lowered by 0.567 K. What is the molar mass of the substance? (Kf of benzene = 5.12 K kg mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 178.9 g mol−1
The depression of freezing point is proportional to the molality of the solute, ΔTf = Kf m. Writing the molality from the masses and solving for the molar mass M2 of the solute (w2 = mass of solute, w1 = mass of solvent in g):
\( \displaystyle M_2=\frac{K_f\times w_2\times1000}{\Delta T_f\times w_1} \)
\( \displaystyle M_2=\frac{5.12\,\mathrm{K\,kg\,mol^{-1}}\times0.44\,\mathrm{g}\times1000\,\mathrm{g\,kg^{-1}}}{0.567\,\mathrm{K}\times22.2\,\mathrm{g}} \)
\( \displaystyle M_2=\frac{2252.8}{12.587}\,\mathrm{g\,mol^{-1}}=178.97\,\mathrm{g\,mol^{-1}} \)
The nearest value is 178.9 g mol−1.