The heat released during mixing of 50 mL of 0.1 $M H_{2} S O_{4}$ with 50 mL of 0.2 $M$ KOH is:

57.3 \times 10^{3} J

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573 J

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5.73 \times 10^{3} J

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57.3 \times 10^{5} J

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Solution

The correct option is B $573 J$
First calculate the milliequivalents of $H_{2} S O_{4}$ .
$M e q . o f H_{2} S O_{4} = 0.1 \times 2 \times 50 = 10$
Then, calculate the milliequivalets of $K O H$ .
$M e q . o f K O H = 0.2 \times 50 = 10$
Now, 1000 milliequivalents of $K O H$ on neutralization liberate 573000 J of heat.
Hence, the heat liberated by the neutralization of 10 milliequivalents of KOH is
$∴ H e a t l i b e r a t e d = \frac{57.3 \times 10^{3} \times 10}{1000} = 573 J$

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The heat released during mixing of 50 mL of 0.1 MH2SO4 with 50 mL of 0.2 M KOH is:

The heat released during mixing of 50 mL of 0.1 $M H_{2} S O_{4}$ with 50 mL of 0.2 $M$ KOH is: