When `50 cm^(3)` of `0.2 N H_(2)SO_(4)` is mixed with `50 cm^(3)` of `1 N KOH`, the heat liberated is

Options

A. 11.46 kJ
B. 57.3 kJ
C. 573 kJ
D. 573J

Answer

D

Explanation

For complete neutralization of strong acid and strong base energy released is 57.32kJ/mol
No. of g eq of `H_(2)SO_(4) = (0.2 xx 50)/(1000) = 10^(-2)`
No. of eq of KOH `= (1)/(1000) xx 50 = 5 xx 10^(-2)`
So `=57.32 xx 10^(-2) = 0.5732kJ = 573.2` Joule