Read following statement(s) carefully and select the right option :
(I) The enthalpy of solution of `CaCl_(2). 6H_(2)O` in a large volume of water is endothermic to the extent of 3.5 kcal/mol. If `DeltaH=-23.2 " kcal"` for the reaction,
`CaCl_(2)(s)+6H_(2)O(l)rarrCaCl_(2).6H_(2)O(s)`
then heat of solution of `CaCl_(2)` (anhydrous) in a large volume of water is - 19.7 kcal/mol
(II) For the reaction `2Cl(g)rarrCl_(2)(g)`, the sign of `DeltaH` and `DeltaS` are negative.

Options

A. Statement I and II both are wrong
B. both are correct
C. only I is correct
D. only II is correct

Explanation

To solve the question, we will analyze both statements provided and determine their correctness step by step.

### Step 1: Analyze Statement I

We are given the following information:

1. The enthalpy of solution of `CaCl2.6H2O` in a large volume of water is endothermic to the extent of 3.5 kcal/mol. This means:
\[ \Delta H_{\text{solution}}(CaCl2.6H2O) = +3.5 \text{ kcal/mol} \]

2. The reaction for the formation of `CaCl2.6H2O` from `CaCl2` and water is given as:
\[ CaCl2(s) + 6H2O(l) \rightarrow CaCl2.6H2O(s) \]
with:
\[ \Delta H = -23.2 \text{ kcal} \]

3. We need to find the heat of solution of anhydrous `CaCl2` in a large volume of water.

### Step 2: Write the Reaction for Anhydrous `CaCl2`

The reaction for the dissolution of anhydrous `CaCl2` in water can be represented as:
\[ CaCl2(s) + nH2O(l) \rightarrow CaCl2.nH2O(s) \]

### Step 3: Apply Hess's Law

According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can combine the two reactions:

1. The dissolution of `CaCl2.6H2O`:
\[ CaCl2.6H2O(s) \rightarrow CaCl2(s) + 6H2O(l) \quad \Delta H = +3.5 \text{ kcal} \]

2. The formation of `CaCl2.6H2O` from `CaCl2` and water:
\[ CaCl2(s) + 6H2O(l) \rightarrow CaCl2.6H2O(s) \quad \Delta H = -23.2 \text{ kcal} \]

When we add these two reactions, we get:
\[ CaCl2.6H2O(s) + 6H2O(l) + CaCl2(s) \rightarrow CaCl2(s) + 6H2O(l) + CaCl2.6H2O(s) \]

### Step 4: Calculate the Total Enthalpy Change

Now, we can calculate the overall enthalpy change:
\[ \Delta H_{\text{total}} = +3.5 \text{ kcal} + (-23.2 \text{ kcal}) = -19.7 \text{ kcal} \]

Thus, the heat of solution of anhydrous `CaCl2` in a large volume of water is:
\[ \Delta H_{\text{solution}}(CaCl2) = -19.7 \text{ kcal/mol} \]

### Conclusion for Statement I

This means Statement I is correct.

### Step 5: Analyze Statement II

For the reaction:
\[ 2Cl(g) \rightarrow Cl2(g) \]

1. **Enthalpy Change (ΔH)**: The formation of a bond (Cl-Cl) releases energy, indicating that ΔH is negative.
2. **Entropy Change (ΔS)**: The number of gaseous moles decreases from 2 moles of Cl to 1 mole of Cl2, indicating a decrease in disorder. Thus, ΔS is also negative.

### Conclusion for Statement II

Both ΔH and ΔS are negative, making Statement II correct.

### Final Answer

Both statements are correct. Therefore, the correct option is **B**.

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