A solution containing `30g` of a nonvolatile solute in exactly `90g` water has a vapour pressure of `21.85 mm Hg` at `25^(@)C`. Further `18g` of water is then added to the solution. The resulting solution has vapour pressure of `22.18 mm Hg` at `25^(@)C`. calculate (a) molar mass of the solute, and (b) vapour pressure of water at `25^(@)C`.

Explanation

Let the vapour pressure of water at `25^(@)C` be `p_(0)` and molecular mass of the solute be m
Using Raoult's law in the following form,
`(p_(0)-p_(s))/(p_(s))=(wM)/(Wm)`
For solution (I), `((p_(0)-21.85))/(22.15)=(30xx18)/(90xxm)`....(i)
For solution (II), `((p_(0)-22.15))/(22.15)=(30xx18)/(108xxm)` ....(ii)
Dividing eq. (i) by eq. (ii),
`((p_(0)-21.85))/(21.85)xx(22.15)/((p_(0)-22.15))=(108)/(90)=(6)/(5)`
`p_(0)=23.87` mm og Hg
Substituting the value of `p_(0)` in eq. (i)
`m =67.9`.