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.

Explanation

`{:(Ca(NO_3)_2hArrCu^(2+)+2NO_3^(-)),((1,0,0,"before dissociation")),((1-alpha,alpha,2alpha,"after dissociation")):}`
`therefore` Total moles at equilibrium `= (1+2alpha)=1+2xx0.7 =2.4`
For `Ca (NO_3)_2: (m_"nor")/(m_"exp") =1+2alpha`, therefore `m_"exp" =(m_"nor")/(1+2xx0.7)=(164)/(2.4)=68.33`
Also at `100^@C , =poverset(@)H_2O= 760mm,w=7 g and w =100 g`
Now `(p^@-p_s)/(p^@)=(7xx18)/(68.33xx100)=0.0184 or 1-(p_s)/(p^@) = 0.0184 "that is " (p_s)/(p^@)= 0.9816`
`therefore p_s=0.9816 xx760 =746.01mm`