Question

For a dilute solution containing $2.5\text{g}$ of a non-volatile non-electrolyte solution in $100\text{g}$ of water, the elevation in boiling point at $1\text{atm}$ pressures is $2^{o}C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure $in\text{mm Hg}$ of the solution is: (Take $k_b=0.76{\text{K kg mol}}^{-1}$)

Answer 1

We know that,
$\Delta T_b=K_b\times m2=0.76\times m\Rightarrow m=\frac{2}{0.76}m=\frac{2.5/M.W.}{100}\times 1000\frac{2}{0.76}=\frac{2.5\times 10}{M.W.}\Rightarrow M.W.=\frac{2.5\times 10}{2}\times 0.76\frac{P_{solvent}^o-P_{solution}}{P_{solvent}^o}=x_{solute}$
$x_{solute}=\text{Mole fraction of solute}$
$P_{solvent}^o=1\text{atm}=760\text{mm Hg}$
$\frac{760-P_{solution}}{760}=\frac{2.5/M.W.}{100/18}760-P_{solultion}=\frac{2.5\times 2}{2.5\times 10\times 0.76}\times \frac{18}{100}\times 760760-P_{solution}=36\Rightarrow P_{solution}=724\text{mm Hg}$