At `25^(@)C`, when 1 mole of `MgSO_(4)` was dissolved in water, the heat evolved was found to be 91.2 kJ. One mole of `MgSO_(4) . 7H_(2)O` on dissolution gives a solution of the same composition accompanied by an absorption of 13.8 kJ. The enthalpy of hydration, i.e., `DeltaH_(h)` for the reaction
`MgSO_(4)(s) + 7H_(2)O(l)rarrMgSO_(4) . 7H_(2)O(s)` :

Options

A. `-105 kJ//"mol"`
B. `-77.4 kJ//"mol"`
C. 105 kJ/mol
D. None of these

Explanation

To find the enthalpy of hydration (ΔH_h) for the reaction:

\[ \text{MgSO}_4(s) + 7\text{H}_2\text{O}(l) \rightarrow \text{MgSO}_4 \cdot 7\text{H}_2\text{O}(s) \]

we will use the given data regarding the dissolution of magnesium sulfate (MgSO₄) and magnesium sulfate heptahydrate (MgSO₄·7H₂O).

### Step-by-step Solution:

1. **Identify the reactions and their enthalpy changes**:
- When 1 mole of MgSO₄ is dissolved in water, the reaction can be represented as:
\[ \text{MgSO}_4(s) + n\text{H}_2\text{O}(l) \rightarrow \text{MgSO}_4 \cdot n\text{H}_2\text{O}(s) \]
The heat evolved for this reaction is given as:
\[ \Delta H_1 = -91.2 \text{ kJ/mol} \]

- When 1 mole of MgSO₄·7H₂O is dissolved in water, the reaction is:
\[ \text{MgSO}_4 \cdot 7\text{H}_2\text{O}(s) \rightarrow \text{MgSO}_4 \cdot n\text{H}_2\text{O}(s) + (n-7)\text{H}_2\text{O}(l) \]
The heat absorbed for this reaction is given as:
\[ \Delta H_2 = +13.8 \text{ kJ/mol} \]

2. **Reverse the second reaction**:
- We need to reverse the second reaction to align it with the first reaction:
\[ \text{MgSO}_4 \cdot n\text{H}_2\text{O}(s) \rightarrow \text{MgSO}_4 \cdot 7\text{H}_2\text{O}(s) + (n-7)\text{H}_2\text{O}(l) \]
The enthalpy change for this reversed reaction becomes:
\[ \Delta H_2' = -13.8 \text{ kJ/mol} \]

3. **Add the two reactions**:
- Now we can add the two reactions together:
\[ \text{MgSO}_4(s) + n\text{H}_2\text{O}(l) \rightarrow \text{MgSO}_4 \cdot n\text{H}_2\text{O}(s) \quad (\Delta H_1) \]
\[ \text{MgSO}_4 \cdot n\text{H}_2\text{O}(s) \rightarrow \text{MgSO}_4 \cdot 7\text{H}_2\text{O}(s) + (n-7)\text{H}_2\text{O}(l) \quad (\Delta H_2') \]
- The overall reaction becomes:
\[ \text{MgSO}_4(s) + 7\text{H}_2\text{O}(l) \rightarrow \text{MgSO}_4 \cdot 7\text{H}_2\text{O}(s) \]

4. **Calculate the total enthalpy change**:
- The total enthalpy change for the overall reaction is:
\[ \Delta H_h = \Delta H_1 + \Delta H_2' = -91.2 \text{ kJ/mol} - 13.8 \text{ kJ/mol} \]
\[ \Delta H_h = -105 \text{ kJ/mol} \]

### Final Answer:
The enthalpy of hydration (ΔH_h) for the reaction is:
\[ \Delta H_h = -105 \text{ kJ/mol} \]