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Stoichiometric Calculations

A 1 g sampl...

Question

A $1$ g sample of $H_{2} O_{2}$ containing $x$ percent $H_{2} O_{2}$ by weight, requires $x$ mL of $K M n O_{4}$ solution for complete oxidation in acidic medium. Calculate the normality of $K M n O_{4}$ solution as nearest integer.

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Solution

The equivalent masses of $H_{2} O_{2}$ and $K M n O_{4}$ are $17$ and $31.6$ respectively.
Let $N$ be the normality of the $K M n O_{4}$ solution.
$x$ mL of $1$ N $K M n O_{4} =$ $x$ mL of $1$ N $H_{2} O_{2} = \frac{x \times N \times 17}{1000} g o f H_{2} O_{2}$ .
$1$ g of the solution contains $\frac{x \times N \times 17}{1000} g H_{2} O_{2}$ .
Also, $1$ g of $H_{2} O_{2}$ contains $= \frac{x}{100} \times 1 g H_{2} O_{2}$ .
Hence, $\frac{17 N \times x}{1000} = \frac{x}{100}$ .
Hence, the normality of the $K M n O_{4}$ solution is $\frac{10}{17} N = 0.588$ .
The nearest integer is $1$ .

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