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Formation of a Differential Equation from a General Solution

100 mL of ozo...

Question

$100$ mL of ozone at STP were passed through $100$ mL of $10$ volume $H_{2} O_{2}$ solution. What is the volume strength of $H_{2} O_{2}$ after the reaction?

9.5

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9.0

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4.75

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4.5

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Solution

The correct option is A $9.5$
$O_{3} \to O_{2} + O$ ...(i)
$H_{2} O_{2} \to H_{2} O + O$ ....(ii)
$O 1 / 2 v o l + O 1 / 2 v o l \to O_{2} 1 v o l$ .....(iii)
From equations (i) and (ii), we inter that $100 m L$ of $O_{3}$ at STP will produce $100 m L$ of molecular $O_{2}$ as such and $100 m L$ of oxygen molecule after reaction with $H_{2} O_{2}$ .
This new volume of $100 m L$ of molecule oxygen after reaction with $H_{2} O_{2}$ is contributed equally by $O_{3}$ and $H_{2} O_{2}$ . Thus, $50 m L$ of oxygen have been contributed by $H_{2} O_{2}$ .
Again, we know
Volume of $H_{2} O_{2} \times$ Volume strength of $H_{2} O_{2}$ = Volume of $O_{2}$ at STP
$∴ 100 m L$ of $^{'} 10 V^{'} H_{2} O_{2} \equiv 1000 m L$ of $O_{2}$ at STP
After utilization of $50 m L$ of $O_{2}$ , according to equation (iii), the balance $(1000 - 50) = 950 m L$ of $O_{2}$ at STP are still retain-able by $100 m L$ of $H_{2} O_{2}$ .
Hence volume strength of $H_{2} O_{2}$ after reaction
$= \frac{Volume of O_{2} a t S T P}{Volume of H_{2} O_{2}} = \frac{950}{100} = 9.5 V$
$∴$ Volume strength $= 9.5$

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