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

Calcium nitrate dissociates as :
`{:(,"",Ca(NO_(3))_(2) hArr,Ca^(2+) + 2NO_(3)^(-)),(,"Initial moles", 1,"0 0"),(,"After dissociation",1-x,"x 2x"):}`
Total no.of moles after dissociation = 1 - x + x+ 2x .
` = 1+2x`
Here `" "x = 70% = 0.7`
No. of moles after dissociation ` 1 + 2 xx 0.7= 2.4`
` i = ("Moles of solute after dissocation")/("Normal moles of solute") = (2.4)/(1) = 2.4`
Now , `(p_(A)^(@) - p_(A))/(p_(A)^(@)) = i x_(B)`
`x_(B) = (n_(B))/(n_(A) + n_(B)) ~~ (n_(B))/(n_(A)) = (w_(B) xx M_(A))/(M_(b) xx w_(A))` (For dilute solution)
`w_(B) = 7.0 g, " " w_(A) = 100g , M_(B) = 164, 18`
`therefore " "w_(B)= (7.0 xx 18)/(164 xx 1000) = 0.00758`
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = 2.4 xx 0.00768`
`" "p_(A)^(@) - p_(A) = 2.4 xx 0.00786 xx 760 = 14.0 mm Hg`
`therefore" "P_(A) = 760- 14.0 = 746 mm Hg`