What is the volume strength of 1.5 N $H_{2} O_{2}$ solution?

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Solution

The volume strength of the 1.5 N $H_{2} O_{2}$ solution:
Calculating the volume strength,
$Strength of 1.5 N H_{2} O_{2} = normality \times equivalent weight (H_{2} O_{2} = 17)$
$Strength of 1.5 N H_{2} O_{2} = 1.5 \times 17$
$Strength of 1.5 N H_{2} O_{2} = 25.5$
Here in this equation,
2 moles of $H_{2} O_{2}$ produce 1 mole of $O_{2}$ .
1 mole of $H_{2} O_{2}$ has weight = $34$
Therefore, mass of 2 moles of $H_{2} O_{2}$ = $68 g / mol$
$68 g / mol$ of $H_{2} O_{2}$ thus producing 1 mole which is exactly equal to $22.4 L$ of oxygen gas at NTP

$So, the volume strength will be, \frac{Vol of O_{2} at NTP}{mass of H_{2} O_{2}} \times strength of H_{2} O_{2}$
$Volume strength = \frac{22.4}{68} \times 25.5 = 8.4 L$
$Hence, the volume strength of 1.5 N H_{2} O_{2} solution is 8.4 liter .$

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$1) I_{2} + H_{2} O_{2} + 2 {OH}^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}$

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