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A solution of a non-volatile solute in w...

A solution of a non-volatile solute in water freezes at `-0.30^(@)C`. The vapour pressure of pure water at `298 K` is `23.51 mm Hg` and `K_(f)` for water is `1.86 degree//molal`. Calculate the vapour pressure of this solution at `298 K`.

Text Solution

Verified by Experts

We have `Delta T = K_(f) xx "molality"`
Also from Raoult's law
`(P^(@)-P_(S))/(P_(S)) = (w xx M)/(m xx W) = (w xx 1000 xx M)/(m xx W xx 1000)`
`(P^(@)-P_(S))/(P_(S)) = "molality" xx (M)/(100)`
`(P^(@)-P_(S))/(P_(S)) = (Delta T)/(K_(f)) xx (M)/(1000)`
Given `P^(@) = 23.51 mm` of `Hg`, `Delta T = 0.3`,
`K_(f) = 1.86 K "molality"^(-1) , M = 18`
`:. (23.51-P_(S))/(P_(S)) = (0.3)/( 1.86) xx (18)/(1000)`
`:. P_(S) = 23.44 mm Hg`
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