One mole of anhydrous salt AB dissolves in water and liberates `15 J mol^(-1)` of heat .The value of `Delta H_(hydration)^°` of AB is -20.05 J `mol^(-1)` .Hence the enthalpy of dissolution of hydrated salt `AB.3H_2O(s)` is

Options

A. `-5.5J mol^(-1)`
B. `5.5J mol^(-1)`
C. 35.5J mol^(-1)
D. -35.5J mol^(-1)

Answer

To solve the problem, we need to find the enthalpy of dissolution of the hydrated salt \( AB \cdot 3H_2O(s) \).

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- The heat released when 1 mole of anhydrous salt \( AB \) dissolves in water is \( 15 \, \text{J/mol} \).
- The enthalpy of hydration \( \Delta H_{\text{hydration}}^° \) of \( AB \) is given as \( -20.05 \, \text{J/mol} \).

2. **Setting Up the Equation**:
- The enthalpy change for the dissolution process can be expressed as:
\[ \Delta H_{\text{dissolution}} = \Delta H_{\text{hydration}} + \Delta H_{\text{dissolution of hydrated salt}} \]

3. **Substituting the Known Values**:
- We know that the heat released during the dissolution of the anhydrous salt is \( -15 \, \text{J/mol} \) (since it is exothermic, we take it as negative).
- Thus, we can write:
\[ -15 = -20.05 + \Delta H_{\text{dissolution of hydrated salt}} \]

4. **Solving for \( \Delta H_{\text{dissolution of hydrated salt}} \)**:
- Rearranging the equation gives:
\[ \Delta H_{\text{dissolution of hydrated salt}} = -15 + 20.05 \]
- Performing the calculation:
\[ \Delta H_{\text{dissolution of hydrated salt}} = 5.05 \, \text{J/mol} \]

5. **Conclusion**:
- The enthalpy of dissolution of the hydrated salt \( AB \cdot 3H_2O(s) \) is \( 5.05 \, \text{J/mol} \).

### Final Answer:
\[ \Delta H_{\text{dissolution of } AB \cdot 3H_2O(s)} = 5.05 \, \text{J/mol} \]