A solution containing `30 g` of a non-volatile solute in exactly `90 g` of water has a vapour pressure of `21.85 mm` of `25^(@)C` . Further `18 g` of water is then added to the solution, the new vapour pressure becomes `22.15 mm` of `Hg` at `25 C`. Calculate the (a) molecular mass of the solute and (b) vapour pressure of water at `25^(@)C`.

Answer

(a) `61.21 g mol^(-1)`, (b) `23.99 mm Hg`

Explanation

`(P^(@) -21.85)/(21.85) = (30xx 18)/(90 xxm)` for I case...I
Now weight of solvent `=90 +18 = 108 g`
`(P^(@) -22.15)/(22.15) = (30 xx 18)/(108 xx m)` for II case...II
`{:( :.,By eq.(i),P^(@)m-21.85m=21.85xx6 = 131.1),(,Byeq.(ii),P^(@)m-22.15m=22.15xx5 = 110.75),(,,ul" + - +"):}`
`:. 0.30 m = 20.35` ltBRgt `:. M = (20.35)/(0.30) =67.83`
`(P^(@) -21.85)/(21.85) = (30 xx18)/(90 xx 67.83)`
`:. P^(@) = 23.78 mm`