Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in $100g$ of water, the elevation in boiling point at $1atm$ pressure is $2℃$. Assuming that the concentration of solute is much lower than the concentration of solvent, the vapour pressure $\left(\text{in mm of Hg}\right)$ of the solution is:

[Take : $K_b=0.76Kkgmo{l}^{-1}$]

Answer 1

Molality $=\frac{△T_b}{K_b}=\frac{2}{0.76}=2.63mol/kg$

and molality of pure water is $55.6mol/kg$

Now,

Lowering invapour pressure $=\frac{P_o-P}{P_o}=\frac{n}{n+N}$

$=\frac{760-P}{760}=\frac{2.63}{55.56}$

On substituting the value and solving it we get the values as $P=724mmHg$.