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Percentage Composition

If x% of ox...

Question

If $x %$ of oxalate ion in a given sample of oxalate salt of which $0.6$ g dissolved in $100$ mL of water required 90 mLof $\frac{M}{100} K M n O_{4}$ for complete oxidation, then the value of $x$ is:

33

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40

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30

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None of these

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Solution

The correct option is A $33$
$M w$ of $C_{2} O_{4}^{2 -} = 88$
$E w = \frac{88}{2} = 44 (n = 2)$
mEq of $C_{2} O_{4}^{2 -} n = 2$ = mEq of $M n O_{4}^{⊝} n = 5$
$\frac{W e i g h t \times 10^{3}}{44} = 90 \times \frac{1}{100} \times 5$
Weight of $C_{2} O_{4}^{2 -} = 0.198 g$
$%$ of $C O_{4}^{2 -} = \frac{0.198 \times 100}{0.6} = 33 %$

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If x% of oxalate ion in a given sample of oxalate salt of which 0.6 g dissolved in 100 mL of water required 90 mLof M100KMnO4 for complete oxidation, then the value of x is:

The correct option is A 33Mw of C2O2−4=88Ew=882=44(n=2)mEq of C2O2−4n=2= mEq of MnO⊝4n=5Weight×10344=90×1100×5Weight of C2O2−4=0.198g% of CO2−4=0.198×1000.6=33%

If $x %$ of oxalate ion in a given sample of oxalate salt of which $0.6$ g dissolved in $100$ mL of water required 90 mLof $\frac{M}{100} K M n O_{4}$ for complete oxidation, then the value of $x$ is: