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Mole Concept and Avogadro's Number

20 mL of H2...

Question

$20$ mL of $H_{2} O_{2}$ solution is reacted with $80$ mL of $0.05 M K M n O_{4}$ in acidic medium. What is the volume strength of $H_{2} O_{2}$ ?

A

2.8

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B

5.6

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C

11.2

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D

None of the above

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Solution

The correct option is B $5.6$
Given:

n factor of $H_{2} O_{2}^{- 2}$ :

When we add 20 ml of $H_{2} O_{2}$ to 80 ml of $K M n O_{4}$ in acidic medium , $H_{2} O_{2}$ is oxidised and $K M n O_{4}$ is reduced.

Reaction is written as follows:
$H_{2} O_{2} + K M n O_{4} \to \frac{1}{2} O_{2} + M n^{2 +}$

In acidic medium, we know that n factor of $K M n O_{4}$ is 5.

Milliequivalent of $H_{2} O_{2}$ = milliequivalent of $K M n O_{4}$

$N o r m a l i t y \times V o l u m e = n f a c t o r \times m o l a r i t y \times v o l u m e$

$N \times 20 = 5 \times 0.05 \times 80$

$N = 1$

By using the formula $N = \frac{V o l u m e s t r e n g t h o f H_{2} O_{2}}{5.6}$
Volume strength of $H_{2} O_{2}$ =5.6

Hence the correct option is B.

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