Question

The vapour pressure of solution of 5g of non electrolyte in 100 g of a water at a particular temperature is $2985N{m}^{-2}$. The vapour pressure of pure water at that temperature is $3000N{m}^{-2}$

Answer 1

Mole fraction of solute, $x_2=\frac{-\Delta p}{p_1^{\ast }}=\frac{3000-2985}{3000}=0.005$

Now, $x_2=\frac{m_2/M_2}{\frac{m_1}{M_1}+\frac{m_2}{M_2}}\approx \frac{m_2/M_2}{m_1/M_1}$

$\therefore 0.005=\frac{5/M_2}{100/18}$

$\Rightarrow M_2=180gmo{l}^{-1}$

$\therefore$ The molar mass of solute is $180gmo{l}^{-1}$ .