The heat of neutralization of LiOH and `H_(2)SO_(4)` at `25^(@)`C is `69.6 KJ mol^(-1)`. Calculate the heat of ionisation of LiOH
will be nearly :-

Options

A. `22.5KJmol^(-1)`.
B. `90 KJ mol^(-1)`.
C. `45KJ mol^(-1)`
D. `33.6KJmol^(-1)`

Explanation

To calculate the heat of ionization of LiOH given the heat of neutralization of LiOH and H2SO4, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Heat of neutralization (ΔH_n) of LiOH and H2SO4 = 69.6 kJ/mol.

2. **Write the Neutralization Reaction**:
The balanced reaction for the neutralization of lithium hydroxide (LiOH) with sulfuric acid (H2SO4) is:
\[ 2 \text{LiOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Li}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

3. **Understand the Heat of Neutralization**:
The heat of neutralization is the heat released when an acid and a base react to form water and salt. For strong acids and bases, this value is typically around -56 kJ/mol for 1 mole of acid and base.

4. **Calculate the Total Heat Released for 2 Moles of LiOH**:
Since the reaction involves 2 moles of LiOH, the total heat released during the neutralization can be calculated as:
\[ \text{Total Heat} = 2 \times (-56 \text{ kJ/mol}) = -112 \text{ kJ} \]

5. **Set Up the Equation for Ionization**:
The heat of ionization (ΔH_i) of LiOH can be calculated using the relationship:
\[ \Delta H_i = \text{Total Heat} + \Delta H_n \]
Substituting the values we have:
\[ \Delta H_i = -112 \text{ kJ} + 69.6 \text{ kJ} = -42.4 \text{ kJ} \]

6. **Calculate the Heat of Ionization per Mole**:
Since the heat calculated is for 2 moles of LiOH, we need to divide by 2 to find the value per mole:
\[ \Delta H_i = \frac{-42.4 \text{ kJ}}{2} = -21.2 \text{ kJ/mol} \]

7. **Final Answer**:
The heat of ionization of LiOH is approximately 21.2 kJ/mol (considering only the magnitude).

### Conclusion:
The final answer is that the heat of ionization of LiOH will be nearly 21.2 kJ/mol.