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Introduction

6.8 g H2O2 ...

Question

6.8 g $H_{2} O_{2}$ is dissolved in 0.2 L solution. Then solution will be labelled as:

22.4 V

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5.6 V

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10 V

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3 V

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Solution

The correct option is B 22.4 V
$6.8 g H_{2} O_{2}$
No. of mole of $H_{2} O_{2} = \frac{6.8}{34} = \frac{1}{5}$
$0.2 L$ contains = $\frac{1}{5}$ mole
$1$ contains $\frac{1}{5 \times 0.2} = 1 m o l e$
$1$ mole of $H_{2} O_{2}$ at $S T P =$ 22.4 V$
Option A

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