Under the same conditions,how many mL of 1 M | vedclass

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(D) The rise in temperature is proportional to the heat evolved during the neutralization reaction. The heat evolved is maximum when the number of equ

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$50, 50$

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(D) The rise in temperature is proportional to the heat evolved during the neutralization reaction. The heat evolved is maximum when the number of equivalents of acid and base are equal.Let the volume of $H_2SO_4$ be $v \ mL$. Then the volume of $KOH$ is $(100 - v) \ mL$.Normality of $H_2SO_4 = 	ext{Molarity} 	imes n	ext{-factor} = 0.5 	imes 2 = 1 \ N$.Normality of $KOH = 	ext{Molarity} 	imes n	ext{-factor} = 1 	imes 1 = 1 \ N$.For maximum neutralization,equivalents of acid = equivalents of base:$v 	imes 1 = (100 - v) 	imes 1$$v = 100 - v$$2v = 100$$v = 50 \ mL$.Thus,volume of $H_2SO_4 = 50 \ mL$ and volume of $KOH = 50 \ mL$.

Under the same conditions,how many $mL$ of $1 \ M \ KOH$ and $0.5 \ M \ H_2SO_4$ solution,respectively,when mixed to form a total volume of $100 \ mL$,produce the maximum rise in temperature?

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