The vapoure pressure of two pure liquids A and B, that from an ideal solution are 100 and 900 torr respectively at temperature T. This liquid solution of A and B is composed of 1 mole of A and 1 mole of B. What will be the pressure, when 1 mole of mixture has been vaporized?

Options

A. `800` torr
B. `500` torr
C. `300` torr
D. None of these

Answer

C

Explanation

Let `N_(B)` mole of `B` present in `1` mole of mixture that has been vaporized. Thus `y_(B)=(eta_(B))/(1)`
Mole fraction of `B` in the remaining liquid phase will be `x_(B)=(1-eta_(B))/(1)`
`x_(B)=(P-P_(T)^(@))/(P_(B)^(@)-P_(T)^(@))` ....(1)
`[because P=P_(T)^(@)+(P_(B)^(@)-P_(T)^(@))x_(B)]`
and `y_(B)=(P_(B))/(P) rArr (P_(B)^(@)x_(B))/(P)` ....(2)
After substitution of values of `x_(B)` and `y_(B)` in (1) and (2)
we get `1-n_(B)=(P-P_(T)^(@))/(P_(B)-P_(T)^(@))` .....(3)
and `n_(B)=((1-n_(B))P_(B)^(@))/(P)` ....(4)
or `n_(B)=(P_(B)^(@))/(P+P_(B))`
so `1-(P_(B)^(@))/(P+P_(B))=(P-P_(T)^(@))/(P_(B)^(@)P_(T)^(@))`
`rArr p=sqrtP_(B)^(@).P_(T)^(@)=sqrt100xx900`
`rArr 300` torr