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JEE Main levelOne or more correct optionsDifficulty 2 of 3eq037

$1.89 g$ of oxalic acid dihydrate $H X_{2} C X_{2} O X_{4} \cdot 2 H X_{2} O$ (molar mass $126 g mo l^{- 1}$ ) is titrated in hot acidic medium with $0.040 M$ $KMnO X_{4}$ . Select the correct statement(s).

AThe oxalic acid sample is $0.030$ equivalent
BThe $KMnO X_{4}$ solution is $0.20 N$
CThe volume of $KMnO X_{4}$ required is $150 mL$
DThe volume of $KMnO X_{4}$ required is $375 mL$

Answer: A, B, C

Equivalent mass of the dihydrate

= 126/2 = 63

, so equivalents

= 1.89/63 = 0.030

.

KMnO X_{4}

(

n = 5

) has

N = 0.040 \times 5 = 0.20

. Volume

= 0.030/0.20 = 150 mL

;

375

mL results only if mole ratio

1 : 1

is wrongly assumed (

0.015/0.040

).

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