Correct Answer - B
Number of moles of `HCl=(MV)/(1000)=(0.01xx25)/(1000)`
`=25xx10^(-5)`
`HCl to H^(+)+Cl^(-)`
`n_(H^(+))=25xx10^(-5)`
Number of moles of `Ca(OH)_(2)=(MV)/(1000)=(0.01xx50)/(1000)=50xx10^(-5)`
`n_(OH^(-))=2xx50xx10^(-5)=10^(-3)`
In the process of neutralisation `25xx10^(-5)` mole `H^(+)` will be completely neutralised.
`thereforeDeltaH=140xx25xx10^(-5)kcal` `=0.035kcal=35cal`