1.95 g of non-volatile and non-electrolyte solute dissolved in 100 g of benzene lowered the freezing point of it by 0.64 K .

The molar mass of the solute (in $\mathrm{g} \mathrm{mol}^{-1}$ )

$$ \left(K_f\left(\mathrm{C}_6 \mathrm{H}_6\right)=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right) $$

At $298 \mathrm{~K}, 0.714$ moles of liquid $A$ is dissolved in 5.555 moles of liquid $B$. The vapour pressure of the resultant solution is 475 torr. The vapour pressure of pure liquid $A$ at the same temperature is 280.7 torr. What is the vapour pressure of pure liquid $B$ in torr?

The mole fractions of glucose and water in aqueous glucose solution are 0.0244 and 0.9756 respectively. What is the weight percentage $(w / w)$ of glucose in this solution?

At $T(\mathrm{~K})$, the vapour pressure of an aqueous solution of a non-volatile solute, whose mole fraction is 0.02 is found to be 34.65 mm Hg . What is the vapour pressure (in mm Hg ) of pure water at the same temperature?