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Normality

Potassium aci...

Question

Potassium acid oxalate $K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} . 4 H_{2} O$ having molar mass 508 g/mol can be oxidized by $M n O_{4}^{-}$ in acidic medium. Calculate the volume of 0.1 M $K M n O_{4}$ reacting in acid solution with one gram of the acid oxalate.

52.80 mL

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15.84 mL

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7.92 mL

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31.68 mL

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Solution

The correct option is D 31.68 mL
$K_{2} C_{2} O_{4} . 3 H_{2} C_{2} O_{4} . 4 H_{2} O; n f = 8$ as each of the 8 carbon atoms in the molecule undergo a +1 change in oxidation state.
$n f$ of permangante ion is 5 as manganese gets reduced from +7 oxidation state to +2 oxidation state.
$Equivalents of oxalate = Equivalents of K M n O_{4}$
$\frac{1}{molecular weight} \times 8 = V \times 0.1 \times 5$
Molecular mass of Potassium acid oxalate = 508 g/mol
$\Rightarrow V = \frac{8}{508 \times 0.1 \times 5} = 0.03168 L$
$\Rightarrow V = 31.68 m l$

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Normality

Standard XII Chemistry

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Potassium acid oxalate K2C2O4.3H2C2O4.4H2O having molar mass 508 g/mol can be oxidized by MnO−4 in acidic medium. Calculate the volume of 0.1 M KMnO4 reacting in acid solution with one gram of the acid oxalate.

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