• # question_answer                 The vapour pressure of a solvent decreased by 10 mm in two columns of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?             A)                   0.8   B)                 0.6          C)                                            0.4          D)                                            0.2

According to Raoult?s law. The relative lowering of vapour pressure is equal to the mole fraction of solute i.e.,                 $\frac{p-{{p}_{s}}}{p}=\frac{n}{n+N}$ $or\frac{\Delta p}{p}=\frac{n}{n+N}$                 $\frac{10}{p}=0.2\,\,\therefore \,\,\,p=50\,\operatorname{m}m$                 For other solution of same solvent                 $\frac{20}{p}=\frac{n}{n+N}$ $or\frac{20}{50}=\frac{n}{n+N}$                 $0.4=\frac{n}{n+N}$                 (mole fraction of solute)                 So, mole fraction of solvent = 1 ? 0.4 = 0.6