The normality of $10$ volume $H_{2} O_{2}$ is nearly?

2.1

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3.4

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1.7

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5.1

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Solution

The correct option is C $1.7$
Normality of $H_{2} O_{2}$ =Molarity $\times$ valency factor of $H_{2} O_{2} = \frac{\frac{x}{11.2}}{1 (L)} \times 2$
where $x =$ volume strength of $H_{2} O_{2}$
$∴$ Normality= $\frac{10}{11.2} \times 2 = 1.7$

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