`100g` of liquid `A(` molar mass `140 g mol ^(-1))` was dissolved in `1000g ` of liquid `B(` molar mass `180g mol^(-1))`. The vapour pressure of pure liquid `B` was found to be `500` torr. Calculate the vapour pressure of pure liquid `A` and its vapour pressure in the solution if the total vapour pressure of the solution is `475 Tor r`

Explanation

Moles of `A = (100)/(140) = 0.714`
Moles of `B = (1000)/(180) = 5.556`
Mole fraction of `A = (0.741)/(0.714+5.556) = 0.114`
Mole fraction of `B = 1 - 0.114 = 0.886`
`P_("total") = p_(A) + p_(B) = p_(A)^(@) x_(A) + p_(B)^(@) x_(B)`
`475 = p_(A)^(@) xx 0.114 + 500 xx 0.886`
`475 = 0.114 p_(A)^(@) + 443`
`0.114 p_(A)^(@) = 32`
`p_(A)^(@) = 280.7 torr`
`therefore` Vap. press of pure A = 280.7 torr .
Vap pressure of A in solution ` = 280.7 xx 0.114` = 32 torr