1-0 g of a sample containing KMnO4 and some inert impurity was dissolved in dilute solution of KOH and volume was made up to 100 mL- To this solution- some ammonia gas was passed where the following redox reaction occured-NH3-KMnO4- N2-MnO2sVolume of N2 measured at N-T-P- was 28 mL- The solution was filtered off to remove MnO2s and a 10 mL portion of the filtrate required 22-5 mL of a 0-01 M oxalate solution to reach the end point- The mass percentage of KMnO4 in the original sample is -

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Question

1.0 g of a sample containing $K M n O_{4}$ and some inert impurity was dissolved in dilute solution of KOH and volume was made up to 100 mL. To this solution, some ammonia gas was passed where the following redox reaction occured:
$N H_{3} + K M n O_{4} \to N_{2} + M n O_{2} (s)$
Volume of $N_{2}$ measured at N.T.P. was 28 mL. The solution was filtered off to remove $M n O_{2} (s)$ and a 10 mL portion of the filtrate required 22.5 mL of a 0.01 M oxalate solution to reach the end point. The mass percentage of $K M n O_{4}$ in the original sample is .

63.2

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63.20

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Solution

Number millimoles of $N_{2}$ produced = $\frac{28}{22.4} = 1.25 m m o l$
The balanced chemical reaction between ammonia and $K M n O_{4}$ is:
$2 N H_{3} + 2 K M n O_{4} \to N_{2} + 2 M n O_{2} + 2 K O H + 2 H_{2} O$
The amount of ammonia passed =2.5 mmol
$\Rightarrow$ ammonia is the limiting reagent as the solution is further titrated against oxalate solution to reduce the left over permanganate.
Number of milli moles of $K M n O_{4}$ reacted with $N H_{3}$ = 2.5
Total number of milli equivalents of oxalate consumed= $22.5 \times 0.01 \times 10 \times 2 = 4.5$
Milli equivalents of $K M n O_{4}$ left unreacted after reaction with ammonia = 4.5
Number of milli moles of $K M n O_{4} = \frac{4.5}{3} = 1.5$
Total number of milli moles of $K M n O_{4} = 2.5 + 1.5 = 4$
$\Rightarrow M a s s o f K M n O_{4} = 4 \times 158 \times 10^{- 3} = 0.632$
Mass % = $\frac{0.632}{1} \times 100 = 63.2$ %

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