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^(@)` and molecular mass of the solute be `Mw_(2)`.
Using Raoult's law in the following form,
`(P^(@) - P_(S)) /P^(S) = (W_(2) Mw_(1)) /( W_(1) Mw_(2))`
For soution (i), `(P^(@) - 21.85) / 21.85 = M/2` ....(i)
For soution (ii), `(P^(@) - 22.15)/(22.15) =( 30 xx 18)/(108 xx Mw_(2))` ....(ii)
Dividing Eq. `(i)` by Eq. `(ii)`, we get
`(P^(@) - 21.85)/(21.85) xx (22.15)/((P^(@)-22.15))=(108)/(90)=(6)/(5)`
`P^(0) = 23.87 mm` of `Hg`
Substituting the value of `P^(@)` in Eq. (i), we get
`Mw_(2) = 67.9`