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The mole fractionof Xin the vapours in equilibrium withhomogenous mixtuerof liquids X andY is 0.42. Thevapour pressure of liquids X and Y at the sametemperatureare 406.5 and 140 torr respectively. Calculate the mole fractionof Xin the solution. |
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Answer» Solution :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 , MOLEFRACTIONS of X and Y are: `y_(X) = (p_(X))/(p_("(total)"))` 0.42 `= (p_(x))/(p_("total")) orp_(X) = 0.42 p_("total")""....(III)` Similarly , `y_(Y) = (p_(y))/(p_("total"))` `0.58 = (p_(y))/(p_("total")) orp_(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` |
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