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

Explanation

To find the heat of hydration of \( \text{CaCl}_2 \) to \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \) using \( \text{H}_2\text{O} (g) \), we can follow these steps:

### Step 1: Write down the given data
1. The enthalpy change for the dissolution of \( \text{CaCl}_2 (s) \) to \( \text{CaCl}_2 (aq) \) is given as:
\[ \Delta H_1 = -75 \, \text{kJ/mol} \]
2. The enthalpy change for the dissolution of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \) to \( \text{CaCl}_2 (aq) \) is given as:
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