The equivalent mass of crystalline oxalic acid (H₂C₂O₄ .2H₂O) is equal to the equivalent weight of:

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HNO₃

CH₃COOH

Na₂CO₃

KOH

answer is B.

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Detailed Solution

$\overset{+ 3}{H_{2} C_{2} O_{4}} \cdot 2 H_{2} O ⟶ {\overset{+ 4}{CO}}_{2}$
n_f =2
$E = \frac{126}{2} = 63$
${HNO}_{3} (n_{f} = 1) E = \frac{63}{1} = 63$

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