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The degree of dissociation of Ca(NO(3))(...

The degree of dissociation of `Ca(NO_(3))_(2)` in a dilute aqueous solution containing `7` g of salt per `100` g of water at `100^(@)`C is `70%`. Calculate the vapour pressure of solution.

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`{:(,Ca(NO_(3))_(2)hArr,Ca^(2+),2NO_(3)^(-),),("Before dissociation",1,0,0,),("After dissociation",1-alpha,alpha,2alpha,):}`
`:.` Total mole at equilibrium `=(1+2alpha) ( :. alpha= 0.7)`
`=(1+2xx0.7)=2.4`
For `Ca(NO_(3))_(2)`: `(m_(N))/(m_(exp))=1+2alpha`
`:. m_(exp)=(m_(N))/((1+2xx0.7))=(164)/(2.4)=68.33`
Also at `100^(@)C, P_(H_(2)o)^(@)=760mm, w= 7 g, W= 100g`
Now `(P^(@)-P_(S))/(P_(S))=(7xx18)/(68.33xx100)=0.0184`
or `(P^(@))/(P_(S))-1= 0.0184`
`:. P_(S)=(760)/(1.0184)= 746.27 mm`
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