Distilled water boils at 373.15 K and freezes at 273.15 K . A solution of glucose in distilled water boils at 373.202 K . What is the freezing point (in K ) of the same solution? (For water, $K_b=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, $K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

An aqueous solution of a non-volatile solute boils at $100.17^{\circ} \mathrm{C}$. The temperature at which this solution will freeze (in ${ }^{\circ} \mathrm{C}$ ) is

$$ \begin{aligned} & \left(K_b\left(\mathrm{H}_2 \mathrm{O}\right)=0.512^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1},\right. \\ & \left.K_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1}\right) \end{aligned} $$

At $50^{\circ} \mathrm{C}$, the vapour pressure of pure benzene is 268 torr. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of 167 torr at the same temperature is (molar mass of benzene $=78 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Liquids $A$ and $B$ form an ideal solution. The vapour pressures of $A$ and $B$ are 50 and 32 mm Hg respectively at 300 K . One mole of liquid $A$ is mixed with 1 mole of liquid $B$. What is the approximate mole fraction of $A$ in vapour phase?