A solution of 394 g of a non-volatile solute in 622 g of water has vapour pressure 30.74 mm Hg at 30 °C. The vapour pressure of pure water at 30 °C is 31.8 mm Hg. The molar mass of the solute is approximately:
A180 g/mol
B180 mg/mol
C342 g/mol
D60 g/mol
Answer & Solution
Correct answer: C. 342 g/mol
$\dfrac{\Delta P}{P_1^0} = \dfrac{W_2 M_1}{M_2 W_1}$ ⇒ $M_2 = \dfrac{W_2 M_1 P_1^0}{\Delta P W_1}$. $\Delta P = 1.06$ mm Hg; $M_1 = 18$ g/mol; $W_1 = 622$ g; $W_2 = 394$ g. $M_2 = (394 \times 18 \times 31.8)/(1.06 \times 622) \approx 342$ g/mol — consistent with sucrose (C₁₂H₂₂O₁₁).
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