Question

Vapour pressure of water at $293K$ is $17.535mmHg$. Calculate the vapour pressure of water at $293K$ when $25g$ of glucose is dissolved in $450g$ of water.

Answer 1

The molar masses of glucose and water are $180$ g/mol and $18$ g/mol respectively.

Vapour pressure of water, p° = 17.535 mm of Hg

Mass of glucose = 25 g

Mass of water = 450 g
Number of moles of glucose $=\frac{25}{180}=0.139$

Number of moles of water $=\frac{450}{18}=25$

Mole fraction of Glucose = $\frac{no.ofmolesofglucose}{no.ofmolesofglucose+no.ofmolesofwater}$

Mole fraction of glucose $=\frac{0.139}{0.139+25}=0.0055$

Vapour pressure lowering is directly proportional to the mole fraction

$\frac{P^0-P}{P^0}=X$

$\frac{17.535-P}{17.535}=0.0055$

$P=17.438$

Hence, the vapour pressure of a solution is $17.438$ mm Hg.