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Types of Redox Reactions

5 litres of a...

Question

$5$ litres of a solution of $H_{2} O_{2}$ with $x$ N strength is diluted to $5.5$ litres. This $5.5$ litres $H_{2} O_{2}$ solution gives $28$ litres $O_{2}$ at NTP. Find the value of $x$ .

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Solution

$5$ litres of a solution of $H_{2} O_{2}$ with $x$ N strength is diluted to $5.5$ litres.
The normality changes to $\frac{5 x}{5.5}$ N.
The normality of the solution and the concentration in g/L are related by the expression, $Normality = \frac{Concentration in g/L}{Equivalent mass}$
$Concentration in g/L = Normality \times Equivalent mass$
The mass of $H_{2} O_{2}$ present in $1$ L of $H_{2} O_{2}$ solution is $\frac{5 x}{5.5} \times 17$ .
The equivalent mass of $H_{2} O_{2}$ is $17$ .
Hence, the mass of $H_{2} O_{2}$ present in $5.5$ L of $H_{2} O_{2}$ solution is $5.5 L \times \frac{5 x}{5.5} \times 17$ ......(1)
At NTP, $68$ g of $H_{2} O_{2}$ gives $22400$ mL of oxygen.
Hence, $28$ litres of oxygen is obtained from $\frac{68}{22400} \times 228000 = 85 g$ of $H_{2} O_{2}$ .
Substitute this in equation (1).
$5.5 L \times \frac{5 x}{5.5} \times 17 = 85$
Hence, $x = 1$ .
Thus, the strength of $H_{2} O_{2}$ solution is $1$ N.

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