`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

`"No. of moles of liquid A (solute)"=("100 g")/("140 g mol"^(-1))=(5)/(7)"mole"`
`"No. of moles of lliquid B (solvent)"=("1000 g")/("180 g mol"^(-1))=(50)/(9)" mole"`
`therefore" Mole fraction of A in the solution "(x_(A))=(5//7)/(5//7+50//9)=(5//7)/(395//63)=(5)/(7)xx(63)/(395)=(45)/(395)=0.114`
`therefore" Mole fraction of B in the solution "(x_(B))=1-0.114=0.886`
`"Also, given "p_(B)^(@)="500 torr"`
`"Applying Raoult's law, "p_(A)=x_(A)p_(A)^(@)=0.114xxp_(A)^(@)" ...(i)"`
`p_(B)=x_(B)p_(B)^(@)=0.886xx500="443 torr"`
`p_("Total")=p_(A)+p_(B)`
`475=0.114p_(A)^(@)+443" or "p_(A)^(@)=(475-443)/(0.114)="280.7 torr"`
Substituting this value in eqn. (i) we get `" "p_(A)=0.114xx"280.7 torr = 32 torr."`