When 0.2 mole of anhydrous `CuSO_(4)` is dissolved in water, the heat evolved is 1.451 kcal. If 0.2moleof `CuSO_(4).5H_(2)O` is dissolved in water, the heat absorbed is 0.264 kcal. Calculate the molar heat of hydration of `CuSO_(4)`

Explanation

To find the molar heat of hydration of `CuSO4`, we will follow these steps:

### Step 1: Determine the heat change for the dissolution of anhydrous `CuSO4`
Given that 0.2 moles of anhydrous `CuSO4` releases 1.451 kcal of heat when dissolved in water, we can calculate the heat change per mole.

\[ \Delta H_1 = \frac{-1.451 \text{ kcal}}{0.2 \text{ moles}} = -7.255 \text{ kcal/mole} \]

### Step 2: Determine the heat change for the dissolution of hydrated `CuSO4.5H2O`
Next, we know that 0.2 moles of `CuSO4.5H2O` absorbs 0.264 kcal of heat when dissolved in water. We can also calculate the heat change per mole for this process.

\[ \Delta H_2 = \frac{0.264 \text{ kcal}}{0.2 \text{ moles}} = 1.32 \text{ kcal/mole} \]

### Step 3: Calculate the molar heat of hydration of `CuSO4`
The molar heat of hydration can be calculated by considering the overall reaction when `CuSO4` is hydrated to form `CuSO4.5H2O`. The reaction can be represented as:

\[ \text{CuSO}_4 (s) + 5 \text{H}_2O (l) \rightarrow \text{CuSO}_4 \cdot 5 \text{H}_2O (s) \]

The heat of the reaction can be calculated as:

\[ \Delta H = \Delta H_1 - \Delta H_2 \]

Substituting the values we calculated:

\[ \Delta H = (-7.255 \text{ kcal/mole}) - (1.32 \text{ kcal/mole}) = -8.575 \text{ kcal/mole} \]

### Conclusion
Thus, the molar heat of hydration of `CuSO4` is:

\[ \Delta H = -8.575 \text{ kcal/mole} \]