One mole of anhydrous `MgCl_(2)` dissolves in water and librates `25 cal//mol` of heat. `Delta H_("hydration")` of `MgCl_(2)=30 cal//mol`. Heat of dissolution of `MgCl.H_(2)O`

Options

A. `+5` cal/mol
B. `-5` cal/mol
C. 55 cal/mol
D. `-55` cal/mol

Explanation

To solve the problem, we need to find the heat of dissolution of `MgCl.H2O`. We will use the given data and apply Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step-by-step Solution:

1. **Identify the reactions and their enthalpy changes:**
- The dissolution of anhydrous `MgCl2` in water:
\[ \text{MgCl}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2 \text{Cl}^-(aq) \quad \Delta H_3 = -25 \text{ cal/mol} \]
- The hydration of `MgCl2`:
\[ \text{Mg}^{2+}(aq) + 2 \text{Cl}^-(aq) \rightarrow \text{MgCl}_2(aq) \quad \Delta H_1 = -30 \text{ cal/mol} \]

2. **Write the desired reaction:**
- The heat of dissolution of `MgCl.H2O` can be represented as:
\[ \text{MgCl}_2(s) + \text{H}_2O(l) \rightarrow \text{MgCl}_2(aq) \quad \Delta H_2 = ? \]

3. **Use Hess's Law:**
- According to Hess's law, we can relate the enthalpy changes:
\[ \Delta H_2 = \Delta H_3 - \Delta H_1 \]
- Substitute the known values:
\[ \Delta H_2 = (-25 \text{ cal/mol}) - (-30 \text{ cal/mol}) \]

4. **Calculate the heat of dissolution:**
- Simplifying the equation:
\[ \Delta H_2 = -25 + 30 = +5 \text{ cal/mol} \]

5. **Conclusion:**
- The heat of dissolution of `MgCl.H2O` is:
\[ \Delta H = +5 \text{ cal/mol} \]

### Final Answer:
The heat of dissolution of `MgCl.H2O` is **+5 cal/mol**.