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Introduction to Oxidation and Reduction

The volume st...

Question

The volume strength of $8.9 M H_{2} O_{2}$ solution calculated at $273 K$ and $1 a t m$ is _____. ( $R = 0.0821 L a t m K^{- 1} m o l^{- 1}$ ) (rounded off to the nearest integer)

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Solution

Volume strength of $H_{2} O_{2}$ is the volume of $O_{2}$ released on decomposition of one volume of $H_{2} O_{2}$ at STP.
$2 H_{2} O_{2} \to 2 H_{2} O + O_{2}$
So, 2 moles of $H_{2} O_{2}$ gives 1 mole of $O_{2}$
Therefore, 1 mole of $H_{2} O_{2}$ will give $11.2 L$ of $O_{2}$ gas.

Volume strength of $H_{2} O_{2}$ at 1 atm, 273 K = M × 11.2
$= 8.9 \times 11.2 = 99.68 = 100$
Hence, answer is 100.

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