1.

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» 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`.


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