2.315 g of a non-volatile solute is dissolved in 49 g of benzene (M = 78 g/mol). The vapour pressure of pure benzene is 640 mm Hg and of the solution is 600 mm Hg. The molar mass of the solute is approximately:
A117 g/mol
B72 g/mol
C38 g/mol
D59 g/mol
Answer & Solution
Correct answer: B. 72 g/mol
$\Delta P = 40$ mm Hg. Using $\dfrac{\Delta P}{P_1^0} = \dfrac{W_2 M_1}{M_2 W_1}$: $M_2 = (W_2 M_1 P_1^0)/(\Delta P W_1) = (2.315 \times 78 \times 640)/(40 \times 49) \approx 72$ g/mol.
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