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 cal
B. `3.5 cal`
C. 10 cal
D. `7.5 cal`

Answer

B

Explanation

no. of milli eq of `Ca(OH)_(2) = 50 xx 0.01 = 0.5`
no. of milli eq of `HCl = 25 xx 0.01 = 0.25` In the process of neutralisation 0.25 milli eq will be completely neutralised
`DeltaH = 0.25 xx 10^(-3) xx 14 xx 10^(3) = 3.5 Cal`