Question

180 g of water contains 6 g urea (Molar mass 60) and 68.4 g sucrose (Molar mass 342) at ${100}^{o}C$. The vapour pressure of solution in mm is_______. (write answer as nearest integer after dividing it by 100)

Answer 1

The number of moles of water $N=\frac{180g}{18g/mol}=10mol$.
The number of moles of solute $=n=\frac{6g}{60g/mol}+\frac{68.4g}{342g/mol}=0.1mol+0.2mol=0.3mol$.
The vapor pressure of pure water is 760 mm Hg.
The expression for the relative lowering of vapor pressure is $\frac{P^0-P}{P^0}=\frac{n}{n+N}$
Substitute values in the above expression.
$\frac{760-P}{760}=\frac{0.3}{10+0.3}=0.029$
$760-p=22.14$
$P=737.86$