At `90^@C`, the vapour pressure of toluene is 400 mm and that of xylene is 150 mm. The mole fraction of toluene in liquid mixture that boil at `90^@C` at a pressure of 0.5 atm is

Options

A. `0.92`
B. `0.86`
C. `0.82`
D. `0.73`

Answer

A

Explanation

The solution boils at 0.5 atm and thus, V.P of mixture `=(760)/(2)mm=380mm because P_(M) = P_(T)^(@).X_(T)+ P_("xylene")^(@).X_("xylene")`
`380 = 400 (X_(T)) + 150 (1-X_(T)) (because X_(T) + X_("xylene")=1)`
`therefore X_("toluene")= 0.92`