At $T(\mathrm{~K})$, the vapour pressure of pure benzene and toluene are 75 and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed to form an ideal solution. If the vapours are in equilibrium with the liquid mixture, the mole fraction of toluene in vapour phase (At.wt. of $\mathrm{C}=12, \mathrm{H}=1$ )

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?