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Balancing Redox Equations

A 3.4 g sampl...

Question

A 3.4 $g$ sample of $H_{2} O_{2}$ solution containing x% by mass requires $x$ ml of a $K M n O_{4}$ solution for complete oxidation under acidic medium. The molarity of $K M n O_{4}$ solution is:

1.0 M

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0.5 M

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0.4 M

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0.2 M

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Solution

The correct option is C 0.4 M
Reaction of $H_{2} O_{2}$ with $K M n O_{4}$ is as
$5 H_{2} O_{2} + 2 K M n O_{4} + 6 H^{+} \to 2 K^{+} + 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}$
Now
100 gm of $H_{2} O_{2}$ soln. contains x gm $H_{2} O_{2}$
So 3.4 gm of soln. will have $(\frac{x}{100}) * 3.4$
mass of $H_{2} O_{2}$ wiil be $(\frac{x}{100}) * 3.4$ gm
The number of moles of $H_{2} O_{2}$ will be
$n_{1} = \frac{g i v e n m a s s}{m o l e c u l a r m a s s}$
` = 0.001x`
Similarly moles of $K M n O_{4}$ will be:
$n_{2}$ = $N o r m a l i t y * v o l u m e$ (in litres)
$= M \times \frac{x}{1000}$
Now $H_{2} O_{2}$ and $K M n O_{4}$ react in the ratio of 5:2 so
$\frac{n_{1}}{5} = \frac{n_{2}}{2}$
$\frac{0.001 x}{5} = \frac{0.001 M x}{2}$
giving Molarity = 0.4 M

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