0-3g of an oxalate was dissolved in 100 mL solution- The solution required 90 mL of N20 KMnO4 for complete oxidation- The - of oxalate ion in salt is-

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Stoichiometric Calculations

0.3g of an ox...

Question

$0.3 g$ of an oxalate was dissolved in $100 m L$ solution. The solution required $90 m L$ of $\frac{N}{20} K M n O_{4}$ 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 D $66 %$
The reaction taking place between manganate and oxalate is as follows:

$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 5 C O_{2} + 8 H_{2} O$

Mn changes from $+ 7$ to $+ 2$ after oxidizing oxalate. Net change in oxidation state $= 5$

C in oxalate changes from $+ 3$ to $+ 4$ after being oxidized. Net change in oxidation state $= 2 \times 1 = 2$

Hence, $1$ equivalent of $K M n O_{4}$ oxidizes $2.5$ equivalences of oxalate.

According to the given data,
Moles of $K M n O_{4} = 0.05 \times \frac{0.09}{5} = 9 \times 10^{- 4}$

Moles of oxalate that $K M n O_{4}$ oxidizes $= 9 \times 10^{- 4} \times 2.5 = = 2.25 \times 10^{- 3}$

Molecular weight of oxalate $(C_{2} O_{4})$ : $2 \times 12 + 4 \times 16 = 88$

Oxalate ion amount $= 2.25 \times 10^{- 3} \times 88 = = 0.198$

Percentage of oxalate ion $= \frac{0.198 \times 100}{0.3} = 66 %$

Hence, the correct option is $B$

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Similar questions

0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mLof $\frac{N}{20} K M n O_{4}$ for complete oxidation. The percentage of oxalate ion in the salt is:

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0.3g of an oxalate was dissolved in 100 mL solution. The solution required 90 mL of N20 KMnO4 for complete oxidation. The % of oxalate ion in salt is:

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