`V.P.` of solute containing `6gm` of non volatile solute in `180gm` of water is `20 "Torr"//mm` of `Hg`. If `1` mole water is further added in to the `V.P.` increase by `0.02`. Torr calculate `V.P.` of pure water `&` molecular of non volatile solute.

Options

A. The molecular weight of solute is `54 g "mol"^(-1)`
B. The vapour pressure of pure water is 20.22 torr
C. Addition of more water in the solution will further raise the vapour pressure of solution
D. The vapour pressure of pure water is 22.22 torr

Answer

A,B,C

Explanation

`(P-20)/20=6/Mxx18/180=6/Mxx1/10`...(1)
`(P-20.02)/20.02=6/Mxx18/198=6/Mxx1/11`...(2)
from (1) & (2) calculation gives `M=54` gm/mole
P=20.22 torr
& on dilution `DeltaP` decreases (lowering) so vapour pressure of solution increases.