Correct Answer - b
`BaCl_(2)(s)+(aq) to BaCl_(2)(aq) , DeltaH =-20.6 kJ`
`BaCl_(2). 2H_(2)O(s)+(aq) to BaCl_(2)(aq),DeltaH=+8.8 kJ` ltbr gt Eq.(i) can be split as
`BaCl_(2)(S)+2H_(2)O(l)to BaCl_(2).2H_(2)O(s),DeltaH=H_(1)`
`BaCl_(2).2H_(2)O(s)+aqto BaCl_(2)(aq),`
`DeltaH=H_(1)+H_(2)+ =-20.6, H_(1) = 8.8 kJ`
` H_(2)=-20.6-8.8 =-29.4 kJ`