Question

$90g$ non-volatile, non-dissociative solute is added to $1746g$ water to form a dilute, ideal solution. The vapour pressure of water has decreased from $300mm$ of $Hgto291mmofHg$ . The molecular weight of solute is.

• A. 90
• B. 60
• C. 30
• D. 15

Answer 1

Answer: (C)

$\frac{\text{P}-{\left(\text{P}\right)}_{\text{s}}}{\text{P}}={\left(\text{X}\right)}_{\text{solute}}\times i\left(i=1\therefore \text{solute is non-ioinsable}\right)$
$\frac{300-291}{300}=\frac{90/M}{\frac{90}{M}+\frac{1746}{18}}$
$\frac{9}{300}=\frac{\frac{90}{M}}{\frac{90}{M}+97}$
$\frac{3\times 90}{M\times 100}+\frac{97\times 3}{100}=\frac{90}{M}$
$\frac{18}{10M}+\frac{97\times 3}{100}=\frac{90}{M}$
$\frac{90}{M}-\frac{18}{10M}-\frac{97\times 3}{100}$
$\frac{900-18}{10M}=\frac{97\times 3}{10}$
$M=\frac{882\times 10}{97\times 3}=30.3\cong 30$