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Standard XII

Chemistry

Concentration Terms

100 ml of giv...

Question

$100$ ml of given $K M n O_{4}$ solution titrates $50$ ml of $0.1$ M oxalic acid. Its normality against alkaline $H_{2} O_{2}$ is:

0.1 N

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0.02 N

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0.06 N

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none of the above

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Solution

The correct option is C $0.06 N$
Milliequivalents of oxalic acid $= 50 \times 0.1 \times 2 = 10$
Medium is acidic $(∴ n = 5$ for $K M n O_{4})$ .
$∴$ Molarity of $K M n O_{4} = \frac{N}{5} = \frac{\frac{10}{100}}{5} = 0.02 M$
In alkaline medium, $n = 3$ for $K M n O_{4}$
Hence, normality $= 0.02 \times 3$ $= 0.06 N$

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100 ml of given KMnO4 solution titrates 50 ml of 0.1 M oxalic acid. Its normality against alkaline H2O2 is:

The correct option is C 0.06NMilliequivalents of oxalic acid =50×0.1×2=10Medium is acidic (∴n=5 for KMnO4).∴ Molarity of KMnO4=N5=101005=0.02MIn alkaline medium, n=3 for KMnO4 Hence, normality =0.02×3 =0.06N

$100$ ml of given $K M n O_{4}$ solution titrates $50$ ml of $0.1$ M oxalic acid. Its normality against alkaline $H_{2} O_{2}$ is: