The integral enthalpies of solution of anhydrous `CuSO_(4)` (s) and hydrated `CuSO_(4).5H_(2)O(s)` are -70 kJ per mol and 10 kJ per mol respectively. Determine the magnitude of enthalpy of hydration of 0.1 mole anhydrous `CuSO_(4)` (s) as
`CuSO_(4)(s)+5H_(2)O(l)rarrCuSO_(4).5H_(2)O(s)`

Answer

To determine the magnitude of the enthalpy of hydration of 0.1 mole of anhydrous CuSO₄ (s), we can follow these steps:

### Step 1: Understand the given data
We know the integral enthalpies of solution for:
- Anhydrous CuSO₄ (s): ΔH₁ = -70 kJ/mol
- Hydrated CuSO₄·5H₂O (s): ΔH₂ = 10 kJ/mol

### Step 2: Write the reaction for hydration
The hydration reaction can be represented as:
\[ \text{CuSO}_4 (s) + 5 \text{H}_2\text{O} (l) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O} (s) \]

### Step 3: Apply Hess's Law
According to Hess's Law, the enthalpy change for the overall reaction can be calculated by considering the enthalpy changes of the individual steps. The enthalpy of hydration (ΔH_hydration) can be calculated as follows:
\[ \Delta H_{\text{hydration}} = \Delta H_2 - \Delta H_1 \]

### Step 4: Substitute the values
Now, substituting the values we have:
\[ \Delta H_{\text{hydration}} = 10 \text{ kJ/mol} - (-70 \text{ kJ/mol}) \]
\[ \Delta H_{\text{hydration}} = 10 \text{ kJ/mol} + 70 \text{ kJ/mol} \]
\[ \Delta H_{\text{hydration}} = 80 \text{ kJ/mol} \]

### Step 5: Calculate for 0.1 mole
To find the enthalpy of hydration for 0.1 mole of anhydrous CuSO₄, we multiply the enthalpy change by the number of moles:
\[ \Delta H_{\text{hydration (0.1 mol)}} = 80 \text{ kJ/mol} \times 0.1 \text{ mol} \]
\[ \Delta H_{\text{hydration (0.1 mol)}} = 8 \text{ kJ} \]

### Final Answer
The magnitude of the enthalpy of hydration of 0.1 mole of anhydrous CuSO₄ is **8 kJ**.

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