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`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. |
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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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