Calculate the enthalpy of hydration of `BaCl_(2)` to `BaCl_(2).2H_(2)O (s)` given that the enthalpy of solution of `BaCl_(2) (s)` is `- 20.6 kJ mol^(-1)` and that of `BaCl_(2).2H_(2)O(s)` is `+ 8.8 KJ mol^(-1)` .

Explanation

Given :
`BaCl_(2)(s) + (aq) to BaCl_(2)(aq) Delta H_(1) = - 20.65 kj mol^(-1)`
`BaCl_(2)2H_(2)O+(aq)toBaCl(2)(aq) Delta H_(2) = 8.8 kj mol^(-1)`
`BaCl_(2)(s) + 2H_(2)O to BaCl_(2)2H_(2)O(s) Delta H = ?`
`Delta H_(1) = Delta H + Delta H_(2)`
`-20.6 kj mol^(-1) = Delta H + 8.8 kj mol^(-1)`
`Delta H = 29.40 kj mol^(-1)`