Question

$90 \, g$ non-volatile, non-dissociative solute is added to $1746 \, g$ water to form a dilute, ideal solution. The vapour pressure of water has decreased from $300 \, mm$ of $Hg \, to \, 291 \, mm \, of \, Hg$ . The molecular weight of solute is.

• A. 90
• B. 60
• C. 30
• D. 15

Answer 1

Answer: (C)

$\frac{\text{P} - \left(\text{P}\right)_{\text{s}}}{\text{P}} = \left(\text{X}\right)_{\text{solute}} \times i \left(i = 1 \therefore \text{solute is non-ioinsable}\right)$
$\frac{300 - 291}{300}=\frac{90 / M}{\frac{90}{M} + \frac{1746}{18}}$
$\frac{9}{300}=\frac{\frac{90}{M}}{\frac{90}{M} + 97}$
$\frac{3 \times 90}{M \times 100}+\frac{97 \times 3}{100}=\frac{90}{M}$
$\frac{18}{10 M}+\frac{97 \times 3}{100}=\frac{90}{M}$
$\frac{90}{M}-\frac{18}{10 M}-\frac{97 \times 3}{100}$
$\frac{900 - 18}{10 M}=\frac{97 \times 3}{10}$
$M=\frac{882 \times 10}{97 \times 3}=30.3\cong30$