$100 m l 0.1 M H_{2} O_{2}$ is mixed with $400 m l 0.5 M H_{2} O_{2}$ . The volume of $0.1 M K M n O_{4}$ (acidic) required to react with $H_{2} O_{2}$ mixture is :

840 ml

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420 ml

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630 ml

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None of these

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Solution

The correct option is A 840 ml
$F i n a l m i l i m o l e s o f H_{2} O_{2} s o l u t i o n$
$= (100) (0.1) + (0.5) (400) = 210 m . m o l$
Now, $m i l i . e q . (H_{2} O_{2}) = m i l i . e q . (K M n O_{4})$
$\Rightarrow (210) (2) = (V) (0.1) (5)$
$\Rightarrow V = 840 m l$

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