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

Let the vapour pressure of X be `p_(1)^(@)` and of Y be `p_(2)^(@)` and `x_(1)` and `x_(2)` be their mole fractions .
Then according to Raoult.s law , the total pressure , p is
` p = p_(1)^(@) . X_(1) + p_(1)^(@) .x_(2)`
In the first solution ,
`x_(1) = (3)/(3+1) = 0 .75 ,x_(2) = (1)/(3+1) = 0.25`
`therefore p_(1)^(@) xx 0.75 + p_(2)^(@) xx 0.25 = 550 mmm Hg`
In the second solution ,
` x_(1) = (4)/(4+1) = 0.80, x_(2) = (1)/(4+1) = 0.20`
` therefore p_(1)^(@) xx 0.80 + p_(2)^(@) xx 0.20 = 560" "........(ii)`
Multiply eq. (i) by 4 and eq.(ii) by 5 we get .
`3p_(1)^(@) + p_(2)^(@) = 2200 mm Hg" "..........(iii)`
`4p_(1)^(@) + p_(2)^(@) = 2800 mm Hg" ".......(iv)`
Subtracting `" " - p_(1)^(@) = - 600` mm Hg
` therefore " "p_(1)^(@) = 600` mm of Hg
Substituting in eq (iii) we get .
`3 xx 600 + p_(2)^(@) = 2200 or p_(2)^(@) = 2200 - 1800 = 400` mm Hg
Vapour pressure of pure component X = 600 mm Hg .
Vapoure pressure of pure componenet Y = 400 mm Hg .