Given `CaCI_(2)(s) +aq rarr CaCI_(2)(aq): DeltaH^(Theta) = 75 kJ mol^(-1)` at `18^(@)C` and `CaCI_(2).6H_(2)O +aq rarr CaCI_(2)(aq), DeltaH^(Theta) 19 kJ mol^(-1)` at `18^(@)C`. Find the heat of hydration of `CaCI_(2)` to `CaCI_(2). 6H_(2)O` by `H_(2)O (g)`. The Heat of vaporisation of water may be taken as `2452 J^(-1)g at 18^(@)C`.

Explanation

`rArr -75 = DeltaH +19 DeltaH = - 75 - 19 =- 94 kJ mol^(-1)`
In the question, `Delta_(hyd)H` from `H_(2)O(g)` is to be calculated, whereas `Delta_(cal)H` above is from `H_(2)O(l)`.
`Delta_(vap)H = 2452 J g^(-1) = 2452 xx 18 J mol^(-1)`
`= 6 xx 2452 xx 18 J mol^(-1)`
`= (6 xx 2452 xx 18)/(10^(3)) kJ mol^(-1)`
`= 264.816`
`DeltaH actual =- 94 -(264.816) =- 358.816 kJ "mole"^(-1)`