The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute solution containing 14g of the salt per 200g of water at `100^(@)C` is 70%. If the vapour pressure of water is 760 mm, calculate the vapour pressure of solution.

Explanation

`Deltap_("theo.")` = Lowering in vapour pressure when there is no dissociation
`=p_o xx(wM)/(Wm)`
(given, `p_0` = 760 mm, w = 14 g, W = 200 g, M = 18, m = 164)
`(760xx14xx18)/(200xx164)=5.84mm`
`"Degree of dissociation"=70/100=0.7`
`Ca(NO_3)_2hArrCa^(2+)+2NO_3-`
( n = 3 )
`(Deltap_(obs.))/(Deltap_("theo."))=("No. of particles after dissociation")/("No. of particles when there is no dissociation")`
`(1+(n-1)alpha)/1=(1+(3-1)xx0.7)/1=2.4`
So, `Deltap_(obs.)=2.4xxDeltap_("theo.")=2.4xx5.84`
= 14.02 mm
`p_0-p_s=Deltap_(0bs.)=14.02`
`p_s=p_0-14.02=760-14.02=745.98mm`