The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute aqueous solution, containing `7.0 g` of the salt per `100 g` of water at `100^(@)C` is `70%`. If the vapour pressure of water at `100^(@)C` is `760 mm`, calculate the vapour pressure of the solution.

Answer

`P_(s)=746.01 mm Hg`

Explanation

Total moles at equilibrium =`1 -alpha+alpha+2alpha=1+2alpha`
`alpha=70%=0.7` (Given)
`i=(1+2alpha)/1=1+2 xx 0.7=2.4`
`Mw_(2)(Ca(NO_(3))_(2))=162 g mol^(-1)`
`Mw_(1)(H_(2)O)=18 g mol^(-1)`
Use the formula,
`(P^(@)-P_(s))/P^(@)=i n_(2)/n_1`
`(760-P_(s))/760=2.4 xx (7xx18)/(164xx100)`
`rArr P_(s) =760 mm Hg`.