Calculate heat of solution of NaCI form the following data:
Hydration energy of `Na^(o+)= - 389 kJ mol^(-1)`
Hydration energy of `CI^(Θ)= - 382 kJ mol^(-1)`
Lattic energy of `NaCI= - 776 kJ mol^(-1)`

Explanation

To calculate the heat of solution of NaCl, we will use the following formula:

\[ \text{Heat of solution} = \text{Hydration energy of Na}^+ + \text{Hydration energy of Cl}^- - \text{Lattice energy of NaCl} \]

### Step 1: Calculate the Hydration Energy of NaCl
The hydration energy of NaCl can be calculated by adding the hydration energies of the individual ions:

\[ \text{Hydration energy of NaCl} = \text{Hydration energy of Na}^+ + \text{Hydration energy of Cl}^- \]

Given values:
- Hydration energy of \( \text{Na}^+ = -389 \, \text{kJ/mol} \)
- Hydration energy of \( \text{Cl}^- = -382 \, \text{kJ/mol} \)

Substituting the values:

\[ \text{Hydration energy of NaCl} = -389 \, \text{kJ/mol} + (-382 \, \text{kJ/mol}) = -771 \, \text{kJ/mol} \]

### Step 2: Use the Lattice Energy
The lattice energy of NaCl is given as:

\[ \text{Lattice energy of NaCl} = -776 \, \text{kJ/mol} \]

### Step 3: Calculate the Heat of Solution
Now, we can substitute the hydration energy and lattice energy into the heat of solution formula:

\[ \text{Heat of solution} = \text{Hydration energy of NaCl} - \text{Lattice energy of NaCl} \]

Substituting the values:

\[ \text{Heat of solution} = -771 \, \text{kJ/mol} - (-776 \, \text{kJ/mol}) \]

This simplifies to:

\[ \text{Heat of solution} = -771 \, \text{kJ/mol} + 776 \, \text{kJ/mol} = 5 \, \text{kJ/mol} \]

### Final Answer
The heat of solution of NaCl is \( 5 \, \text{kJ/mol} \).

---