Question

Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` fo `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm Hg`. Determine the vapour pressure of `A` and `B` in their pure states.

Answer 1

Correct Answer - `P_(A)^(@)=400 mm Hg`;``P_(B)^(@)=600 mm Hg`
Given,
Total vapour pressure of solution =`550 mm` of `Hg`
Moles of `A`=`1`
Moles of `B`=`3`
Total number of moles =`3+1`=`4`
Mole of fraction of `A`=`1/4`
Mole of fraction of `B`=`3/4`
Let the vapour pressure of `A` =`p_(1)^(@)`
and the vapour pressure of `B` =`p_(2)^(@)`
Total `VP` =`VP` of `A` `xx` Mole fraction of `A` + `VP` of `B` `xx` Mole fraction of `B` ....(i)
`550=P_(1)^(@) xx 1/4+P_(2)^(@) xx 3/4`
`4 xx 550 =P_(1)^(@) + 3p_(2)^(@)` ......(ii)
On adding `1 mol` of `B`
Moles of `B` becomes =`3+1=4`
Moles of `A`=`1`
Total number of moles =`4+1=5`
Mole fraction of `A`=`1/5`
Mole fraction of `B`=`4/5`
and `VP` becomes =`550 +10 =560 mm Hg`
From Eqs. (i)
`560=p_(1)^(@) xx 1/5+p_(2)^(@) xx 4/5`
`560 xx 5 =p_(1)^(@) + 4p_(2)^(@)` .....(iii)
From Eqs. (ii) and (iii), we get
`p_(1)^(@)=400 mm Hg`
`p_(2)^(@)=600 mm Hg`
Thus, vapour pressure of pure `A` =`400 mm Hg`
vapour pressure of pure `B` =`600 mm Hg`