Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`

Options

A. `14 kcal`
B. `35 cal`
C. `10 cal`
D. `7.5 cal`

Explanation

To calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)₂` reacts with `25 mL` of `0.01 M HCl`, we will follow these steps:

### Step 1: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in L)} \]
Given:
...