A) 11.46 kJ
B) 57.3 kJ
C) 573 kJ
D) 573 J
Correct Answer: D
Solution :
The number of gram-equivalents of \[{{H}_{2}}S{{O}_{4}}=0.2\times \frac{50}{1000}=1.0\times {{10}^{-2}}\] The number of gram-equivalents of \[KOH=1\times \frac{50}{1000}\] \[=5\times {{10}^{-2}}\] We know that 57.3 kj heat is evolved when one gram-equivalent of strong acid (like\[({{H}_{2}}S{{O}_{4}})\] and one gram-equivalent of strong base (like KOH) are neutralised. Here \[1.0\times {{10}^{-2}}\] gram-equivalent of \[{{H}_{2}}S{{O}_{4}}\] is neutralised by \[1.0\times {{10}^{-2}}\] gram-equivalent of KOH hence the heat evolved will be \[=57.3\,kj\times 1.0\times {{10}^{-2}}\] \[=57.3\,kj\times 1.0\times {{10}^{-2}}\] = 573 j
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