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Percentage Composition

0.3 g of an o...

Question

$0.3 g$ of an oxalate salt was dissolved in $100 m L$ solution. The solution required $90 m L$ of $N / 20 K M n O_{4}$ for complete oxidation.The % of oxalate ion in salt is:

33 %

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66 %

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70 %

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40 %

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Solution

The correct option is B $66 %$
Reaction taking between potassium permanganate and axalate-
$M n O_{4}^{-} + C_{2} O_{4}^{2 -} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O$
Hence equivalent of $K M n O_{4}$ oxidizes $2.5$ equivalences of axalate
Moles of $K M n O_{4} = 0.05 \times \frac{0.09}{5} = 9 \times 10^{- 4} . . . . . (1)$
Moles of oxalate that $M K n O_{4}$ axidizes $= 9 \times 10^{- 4} \times 2.5$
$= 22.5 \times 10^{- 4}$ (from $(1)$ ) $. . . . . . (2)$
Molecolar weight of oxalate $= 2 \times 12 + 4 \times 16 = 88$
( $∵ C_{2} O_{4}$ , where $C = 12$ & $O = 16$ ) $. . . . . . (3)$
Oxalate ion amount $= 22.5 \times 10^{- 4} \times 88 = 0.198$ from $(2)$ & $(3)$
Percentage of oxalate ion $= \frac{0.198 \times 100}{0.3} = 66 %$
Ans $B) 66 %$

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Similar questions

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0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL of N/20 KMnO4 for complete oxidation.The % of oxalate ion in salt is:

$0.3 g$ of an oxalate salt was dissolved in $100 m L$ solution. The solution required $90 m L$ of $N / 20 K M n O_{4}$ for complete oxidation.The % of oxalate ion in salt is: