The amount of heat released when `20 ml`,`0.5 M` `NaOH`is mixed with `100 ml`,`0.1 M` `H_(2)SO_(4)` is `xKJ`. The heat of neutralization will be `:-`
A. `-100x`
B. `-50x`
C. `+100x`
D. `+50x`


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Correct Answer - A
Number of moles of `NaOH=(MV)/(1000)=(0.5xx20)/(1000)=0.01`
Number of moles of `HCl=(MV)/(1000)=(0.1xx100)/(1000)=0.01`
Heat of neutralization `=(-x)/(0.01)=-100x`

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