1.

`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.

Answer» `(P^(0)-P_(s))/(P_(s))=(w)/(m)xx(M)/(W)`
`rArr (P^(0)-20)/(20)=(6)/(m)xx(18)/(180)`
`(P^(@)-20.02)/(20.02)=(6)/(m)xx(18)/(198)`
`rArr P^(@)=20.22` Torr.
`m=54 gm//mol`.


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