One moles of anhydrous `AB` dissolves in water and liberates `21.0 J mol^(-1)` of heat. The valueof `DeltaH^(Theta)` (hydration) of `AB` is `-29.4 J mol^(-1)`. The heat of dissolution of hydrated salt `AB.2H_(2)O(s)` is

Options

A. `50.4 J mol^(-1)`
B. `8.4 J mol^(-1)`
C. `- 50.4 J mol^(-1)`
D. `-8.4 J mol^(-1)`

Answer

B

Explanation

`(i)AB(s)+aqrarrAB(aq),DeltaH = - 21J`
(ii)` AB(s)+xH_(2)O(s)rarrAB.H_(2)O(s),DeltaH=-29.4J`
Aim :` AB(s)+xH_(2)O(s)+aqrarrAB(aq),DeltaH=?`
Eqn. (i0 is decomposed as -
`AB(s)+xH_(2)O(s)rarrAB.H_(2)O(s),DeltaH=DeltaH_(1)`..(a)
`AB.xH_(2)O(s)+aqrarrA(aq),DeltaH=DeltaH_(2)..(b)`
`DeltaH_(1)DeltaH_(2)= - 21 ` `-29.4+DeltaH_(2) = - 21 "or" DeltaH_(2) = 8.4` J `mol^(-1)`