Question

Vapour pressure in mm $Hg$ of $0.1$ mole of urea in 180 g of water at ${25}^{\circ }C$ is (The vapour pressure of water at ${25}^{\circ }C$ is $24$ mm $Hg$)

• A. $2.376$
• B. $20.76$
• C. $23.76$
• D. $24.76$

Answer 1

The correct option is C $23.76$

The molar mass of water is 18 g/mol
The number of moles of water is $\frac{180}{18}=10mol$
The mole fraction of urea is

$X=\frac{0.1}{0.1+10}=0.0099$

The relative lowering in the vapour pressure of solution is equal to the mole fraction of solute
$\frac{p^o-p}{p^o}=X$

$\frac{24-p}{24}=0.0099$

$24-p=0.2376$

$p=23.76mmHg$