`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

Calculation of vapour pressure of pure luquid `A(P_(A)^(@))`.
For an ideal solution: `p=p_(A)^(@)X_(A)+P_(B)^(@)x_(B)`
No. of moles of liquid A `(n_(A))=W_(A)/M_(A)=((100))/((140g mol^(-1)))=0.7143 mol`
No. of moles of liquid B `(n_(B))=W_(B)/M_(B)=((1000g))/((180g mol^(-1)))=5.5556 mol`
Mole fractin of A `(A_(A)) =n_(A)/(n_(A)+n_(B))`
`=((0.7143 mol))/((0.7143 mol + 5.5556 mol))=(0.7143 mol)/(6.2699)=0.1139`
Mole fration of `B (x_(B))=1-00.1139=0.8861`
Vapour pressure of pure liuid `B(_(B)^(@))=500 torr`.
Toral vapour pressure solution `(P)=475 torr.`
The valur of `P_(A)^(@)` can be calculated as follows :
475 torr `P_(A)^(@)xx0.1139+(500 torr)xx(0.8861)`
=`P_(A)^(@)xx0.1139=443.05 torr`
`P_(A)^(@)xx0.1139=475 torr-443.05 torr 31.95 torr`
`P_(A)^(@)=((31.95 torr))/(0.1139)=280.5 torr`
calculation of vapour pressure of A in the solution i.e., `P_(A)`
According to Raoult's law, `P_(A)=P_(A)^(@)x_(A)`
=(280.5 torr)xx0.1139=32.0 torr