Two liquids X and Y one mixing form an ideal solution. At `30^(@)C` the vapour pressure of the solution containing 3 moles of X and 1 mole Y is 550 mm Hg. But when 4 moles of X and 1 mole of Y are mixe, the vapour pressur of the solution thus formed is 560 mm Hg. What will be the vapour presure of pure and Pure Y at this temerature ?

Explanation

Suppose the vapour pressure of pure X is `p_(X)^(@)` and that of pure Y is `p_(Y)^(@)`.
In the first case, mole fraction of `X=(3)/(4)=0.75` and mole fraction of `Y=0.25.`
By Raoult's law,`" "P_("total")=p_(X)+p_(Y)=x_(X)p_(X)^(@)+x_(Y)p_(Y)^(@)" "therefore" "550=0.70p_(X)^(@)+p_(Y)^(@)" ...(i)"`
In the second case, mole fraction of `X=(4)/(5)=0.80` and mole fraction of Y = 0.20
By Raoult's law, `" "560=0.80p_(X)^(@)+0.20p_(Y)^(@)" ...(ii)"`
From eqn. (i),`" "2200=3p_(X)^(@)+p_(Y)^(@)" ...(iii)"`
From eqn. (ii),`" "2800 = 4p_(X)^(@)+p_(Y)^(@)" ...(iv)"`
Subtracting eqn. (iii) from eqn. (iv), we get, `" "p_(X)^(@)=600mm`
Substituting in (iii), we get, `p_(Y)^(@)=2200-3xx600=400mm`