0-3 g of an oxalate salt was dissolved in 100 mL solution- The solution required 90 mLof N -20 KM nO 4 for complete oxidation- The percentage of oxalate ion in the salt is-A- 66 -B- 70 -C- 40 -D- 33 -

The correct option is C 40 Number of mili equivalent of KMnO_4 = 90 ×1/20=4.5 m equivalents. and number of mili equivalents of oxalate ions = number of mili eq ...

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Byju's Answer

Standard XII

Chemistry

Percentage Composition

0.3 g of an o...

Question

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:

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 C. 66%
Number of mili equivalent of $K M n O_{4} = 90 \times \frac{1}{20} = 4.5 m$ equivalents.

and number of mili equivalents of oxalate ions = number of mili equivalents of $K M n O_{4} = 4.5 m$ equivalents

$∵$ Number of equivalents $= \frac{Mass}{equivalent mass}$

& equivalent mass of oxalate ion $= \frac{88}{2} = 44$

$∴$ Mass of oxalate $= 44 \times 4.5 \times 10^{- 3} g r a m s$

$= 198 \times 10^{- 3} g$

= 0.198 g

$∴$ percentage in salt oxalate $= \frac{0.198}{0.3} \times 100 = 66 %$

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0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mLof N20KMnO4 for complete oxidation. The percentage of oxalate ion in the salt is:

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: