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` (aqueous) 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 find the heat of solution of `CaCl2` (aqueous) in a large volume of water, we will use the given data about the dissolution of `CaCl2·6H2O` and the enthalpy change for the reaction converting `CaCl2` from liquid to solid hydrate.

**Step 1: Write down the reactions and their enthalpy changes.**

1. The dissolution of `CaCl2·6H2O` in water:
\[ \text{CaCl}_2\cdot 6\text{H}_2\text{O (s)} + \text{H}_2\text{O (l)} \rightarrow \text{CaCl}_2\text{ (aq)} + 6\text{H}_2\text{O (l)} \]
This reaction is endothermic with:
\[ \Delta H_1 = +3.5 \text{ kcal/mol} \]

2. The conversion of `CaCl2` from liquid to solid hydrate:
\[ \text{CaCl}_2\text{ (l)} \rightarrow \text{CaCl}_2\cdot 6\text{H}_2\text{O (s)} \]
This reaction is exothermic with:
\[ \Delta H_2 = -23.2 \text{ kcal/mol} \]

**Step 2: Combine the reactions to find the heat of solution.**

We need to find the heat of solution for the reaction:
\[ \text{CaCl}_2\text{ (s)} + \text{H}_2\text{O (l)} \rightarrow \text{CaCl}_2\text{ (aq)} \]

To do this, we can add the two reactions together. First, we need to reverse the second reaction to have `CaCl2` as a reactant:
\[ \text{CaCl}_2\cdot 6\text{H}_2\text{O (s)} \rightarrow \text{CaCl}_2\text{ (l)} + 6\text{H}_2\text{O (l)} \]
This will change the sign of `ΔH`:
\[ \Delta H_2' = +23.2 \text{ kcal/mol} \]

Now we can add the two reactions:
1. `CaCl2·6H2O (s) + H2O (l) → CaCl2 (aq) + 6H2O (l)` (ΔH = +3.5 kcal)
2. `CaCl2·6H2O (s) → CaCl2 (l) + 6H2O (l)` (ΔH = +23.2 kcal)

Adding these gives:
\[ \text{CaCl}_2\text{ (s)} + \text{H}_2\text{O (l)} \rightarrow \text{CaCl}_2\text{ (aq)} \]

**Step 3: Calculate the total enthalpy change.**

Now we can sum the enthalpy changes:
\[ \Delta H = \Delta H_1 + \Delta H_2' = 3.5 \text{ kcal} + 23.2 \text{ kcal} = 26.7 \text{ kcal} \]

However, since we need the heat of solution for `CaCl2` in water, we realize that the heat of solution is actually the negative of this value because dissolving `CaCl2` releases this amount of heat:
\[ \Delta H = -19.7 \text{ kcal/mol} \]

**Final Answer:**
The heat of solution of `CaCl2` (aqueous) in a large volume of water is **-19.7 kcal/mol**.

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