A) 11.46 kJ
B) 57.3 kJ
C) 573 kJ
D) 573 J
Correct Answer: D
Solution :
For complete neutralization of strong acid and strong base energy released is \[57.32\,KJ/mol\] No. of mole of \[{{H}_{2}}S{{O}_{4}}=\frac{0.2\times 50}{1000}={{10}^{-2}}\] No. of mole of \[KOH=\frac{1}{1000}\times 50=5\times {{10}^{-2}}\] So \[=57.32\times {{10}^{-2}}=0.5732\,KJ\] \[=573.2\,\,Joule\].You need to login to perform this action.
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