The dissolution of `CaCl_(2)cdot 6H_(2)O` in large volume of water is endothermic to the extent of `3.5 kcal mol^(-1)` . For the reaction `CaCl_(2)(l) rarr CaCl_(2) 6H_(2)O(s) " "Delta H = -23.2 kcal`
Hence, heat of solution of `CaCl_2` (anhydrous) in a large volume of water is:

Options

A. `26.7 kcal `
B. `-26.7 kcal`
C. `19.7 kcal`
D. `-19.7 kcal`

Explanation

To solve the problem, we need to determine the heat of solution of anhydrous \( \text{CaCl}_2 \) in a large volume of water. We are given two reactions with their respective enthalpy changes.

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Enthalpies:**
- The dissolution of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \) in water is endothermic with \( \Delta H_1 = +3.5 \, \text{kcal/mol} \).
- The reaction of converting \( \text{CaCl}_2 \) solid to \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \) solid has \( \Delta H_2 = -23.2 \, \text{kcal} \).

2. **Write the Reactions:**
- For the dissolution of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \):
\[ \text{CaCl}_2 \cdot 6\text{H}_2\text{O} (s) + \text{H}_2\text{O} (l) \rightarrow \text{CaCl}_2 (aq) \]
- For the formation of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \):
\[ \text{CaCl}_2 (s) \rightarrow \text{CaCl}_2 \cdot 6\text{H}_2\text{O} (s) \]

3. **Combine the Reactions:**
- To find the heat of solution of anhydrous \( \text{CaCl}_2 \), we need to add the two reactions:
- Start with the dissolution of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \) and then reverse the formation of \( \text{CaCl}_2 \cdot 6\text{H}_2\text{O} \):
\[ \text{CaCl}_2 (s) \rightarrow \text{CaCl}_2 \cdot 6\text{H}_2\text{O} (s) \quad (\Delta H = -23.2 \, \text{kcal}) \]
\[ \text{CaCl}_2 \cdot 6\text{H}_2\text{O} (s) + \text{H}_2\text{O} (l) \rightarrow \text{CaCl}_2 (aq) \quad (\Delta H = +3.5 \, \text{kcal}) \]

4. **Calculate the Overall Enthalpy Change:**
- The overall reaction can be written as:
\[ \text{CaCl}_2 (s) + \text{H}_2\text{O} (l) \rightarrow \text{CaCl}_2 (aq) \]
- The enthalpy change for this overall reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2 = 3.5 \, \text{kcal} + (-23.2 \, \text{kcal}) = -19.7 \, \text{kcal} \]

5. **Conclusion:**
- The heat of solution of anhydrous \( \text{CaCl}_2 \) in a large volume of water is \( \Delta H = -19.7 \, \text{kcal} \).

### Final Answer:
The heat of solution of \( \text{CaCl}_2 \) (anhydrous) in a large volume of water is \( -19.7 \, \text{kcal} \).

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