The mole fraction of X in the vapours in equilibrium with homogenous mixtuer of liquids X and Y is 0.42. The vapour pressure of liquids X and Y at the same temperature are 406.5 and 140 torr respectively. Calculate the mole fraction of X in the solution.

Explanation

According the Raoult.s law,
`p_(X) = p_(X)^(@) xx x_(X)" "…..(i)`
`p_(Y) = p_(Y)^(@) xx x_(Y) = p_(Y)^(@) (1-x)" "…….(ii)`
In vapour state , mole fractions of X and Y are :
`y_(X) = (p_(X))/(p_("(total)"))`
0.42 `= (p_(x))/(p_("total")) or p_(X) = 0.42 p_("total")" "....(iii)`
Similarly , `y_(Y) = (p_(y))/(p_("total"))`
`0.58 = (p_(y))/(p_("total")) or p_(Y) = 0.58 p_("total") " ".......(iv)`
Dividing eq.(iii) by eq,(iv).
or ` " "(p_(y))/(p_("total")) = (0.42)/(0.58) = 0.724`
From eq.(i) ,(ii) and eq.(v) ,
`(p_(x))/(p_(y)) = (p_(x)^(@) xx x_(x))/(p_(y)^(@) xx (1-x_(x))) = 0.724`
` or (406.5x)/(140(1-x)) = 0.724`
or `" " 406.5 x_(x) = 101.36 - 101.36 x_(X)`
`507.86 x_(X) = 101.36`
`therefore " " x _(X) = 0.20`