Question

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

Answer 1

Correct Answer - 2
`m.eq.` of base `=50xx0.01xx2=1`
`m.eq.` of acid `=25xx0.01" "0.25`
`:. Me.eq.` of base reacted `=0.25`
`:.` energy released `=(140)/(1000)xx0.25Kcal=(140xx0.25xx1000)/(1000)cal=35cal`.