100g of liquid A (molar mass 140g `"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 torr.

Answer

`P^(@)=32 t o r r`

Explanation

Since the amount of liquid `B` is more, so liquid `B` is solvent and liquid `A` is solute.
`:. N_(B)("solvent")=(1000g)/(180g mol^(-1))=5.555`
`n_(A)("solvent")=(100g)/(140g mol^(-1))=0.714`
`CHMi_(B)=(5.555)/(5.555+0.714)=0.886`
`CHMi_(A)=1-0.886=0.114`

Also given, `p_(B)^(@)=500t o r r`
Using Raoult's law
`p_(A)=CHMi_(A)p_(A)^(@)=0.114xxp_(A)^(@) ....(i)`
`p_(B)=CHMi_(B)p_(B)^(@)=0.886xx500=443t o rr`
`p_(t otal)=p_(A)+p_(B)` ltBRgt `475=0.114p_(A)^(@)+443`
or `p_(A)^(@)=(475-443)/(0.114)=280.7 t o r r`
Substituting the value of `p_(S)^(@)` in `Eq. (i)`, we get
`p_(A)=0.114xx280.7=32` torr .