Question

How much urea (molar mass $60g/mol$) should be dissolved in $50g$ of water so that its vapour pressure at room temperature is reduced by $25\%$.[Write up to 1 decimal place]

Answer 1

We have $\frac{P_1^o-P_1}{P_1^o}=\frac{W_1\times M_2}{W_2\times M_1}⟶\left(1\right)$

where, $P_1^o$= Vapour pressure of pure solvent= $x$

$P^o$= Vapour pressure of solution of urea

$\Rightarrow P_1^o-P_1=25$% of $x=0.25x$

$M_1$= Mass of water= $18.02g/mol,W_1=50g$

$M_2=60g/mol$

$\therefore \left(1\right)\Rightarrow \frac{0.25x}{x}=\frac{W_2\times 18.02}{50\times 60}$

$\Rightarrow W_2=\frac{0.25\times 3000}{18.02}=41.62g\simeq 41.7g$