100g of liquid A(molar mass `140"g mol"^(-1)`) was dissolved in 1000g of liquid B(molar mass `180"g 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 torr.

Explanation

Calculation of vapour pressure of pure liquid A `(P_(A)^(0))`
Number of moles of liquid A
`n_(A)=(w)/(m)=(100)/(140)=0.7143`
Number of moles of liquid B
`n_(B)=(w)/(m)=(1000)/(180)=5.5556`
Mole fraction of A
`X_(A)=(n_(A))/(n_(A)+n_(B))=(0.7143)/(0.7143+5.5556)=(0.7143)/(6.2699)`
`=0.1139`
Mole fraction of `B(X_(B))=1-0.1139=0.8861`
Vapour pressure of liquid `(P_(B)^(0))=500" torr"`
Total vapour pressure of solution (P) = 475 torr
AC Cording to Raoult's law
`P=P_(A)^(0)X_(A)+P_(B)^(0)X_(B)`
`475=P_(A)^(0)(0.1139)+500(0.8861)`
`475=P_(A)^(0)(0.1139)+443.05`
`P_(A)^(0)=(475-443.05)/(0.1139)=(31.95)/(0.1139)=280.5" torr"`
Calculation of vapour pressure of A in the solution `(P_(A))`
`P_(A)=P_(A)^(0)X_(A)=280.5xx0.1139`
`P_(A)=32` torr.