Question

1 mole $A_X{B}_Y$ when added to 1.08 kg of water, lowers its vapour pressure by 35 torr to 700 torr. If further 1184 g of methanol is added, then the vapour pressure increase by 37 torr. Consider only few vapors are formed, and $A_X{B}_Y$ is a non volatile solute. What is the value of a if the vapour pressure of pure methanol is $a\times {10}^2$ torr.

Answer 1

When 1 mole of $A_X{B}_Y$ is added to 1.08 kg of water, the vapor pressure is lowered from 735 torr to 700 torr. When 1184 g of methanol is added, the vapor pressure is increased from 735 torr to $735+37=772torr=7.72\times {10}^{2}torr=a\times {10}^{2}torr$.
Hence, $a=7.72\approx 8$.