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`.

Options

A. `74.2`
B. `75.6`
C. `67.83`
D. `78.7`

Answer

C

Explanation

We have,`(P_(0)-21.85)/(21.85)=(30xx18)/(90xxm)`, for`I`case….(i)
wt.of solvent `=90+18=108 gm`
`(P^(0)-22.15)/(22.15)=(30xx18)/(108xxm)`,for `II` case….(ii)
By eq.`(1)P_(m)^(0)-21.85m=21.85xx3=131.1`
by eq. (2) `P^(0)m-22.15m=22.15xx5=110.75`
`0.30m=20.35`
`m=(20.35)/(0.30)=67.83`