Lattice energy of `NaCl_((s))` is `-788kJ mol^(-1)` and enthalpy of hydration is `-784kJ mol^(-1)`. Calculate the heat of solution of `NaCl_((s))`.

Options

A. `4 kJ mol^(-1)`
B. `-4 kJ mol^(-1)`
C. `-1572 kJ mol^(-1)`
D. `1572 kJ mol^(-1)`

Explanation

To calculate the heat of solution (enthalpy of solution) of NaCl, we can use the relationship between lattice energy, enthalpy of hydration, and enthalpy of solution. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- **Lattice Energy (U)**: The energy required to separate one mole of an ionic solid into gaseous ions. For NaCl, it is given as `-788 kJ/mol`.
- **Enthalpy of Hydration (ΔH_hyd)**: The energy released when one mole of gaseous ions is hydrated. For NaCl, it is given as `-784 kJ/mol`.
- **Enthalpy of Solution (ΔH_sol)**: The heat change when one mole of solute dissolves in a solvent.

2. **Write the Relationship**:
The relationship between these quantities can be expressed as:
\[ \text{Lattice Energy} = \text{Enthalpy of Hydration} + \text{Enthalpy of Solution} \]
Mathematically, this can be rearranged to find the enthalpy of solution:
\[ \Delta H_{sol} = \text{Lattice Energy} - \text{Enthalpy of Hydration} \]

3. **Substitute the Given Values**:
- Lattice Energy (U) = `-788 kJ/mol`
- Enthalpy of Hydration (ΔH_hyd) = `-784 kJ/mol`
Substitute these values into the equation:
\[ \Delta H_{sol} = -788 \, \text{kJ/mol} - (-784 \, \text{kJ/mol}) \]

4. **Calculate the Enthalpy of Solution**:
\[ \Delta H_{sol} = -788 + 784 = -4 \, \text{kJ/mol} \]

5. **Conclusion**:
The heat of solution (enthalpy of solution) of NaCl is `-4 kJ/mol`.

### Final Answer:
The heat of solution of NaCl is `-4 kJ/mol`.