Question

Molecular weight of oxalic acid is 126. The weight of oxalic acid required to neutralise 100 cc of normal solution of $\text{N}\text{a}\text{O}\text{H}$ is

• A. $6.3 \, \text{g}\text{m}$
• B. $126 \, \text{g}\text{m}$
• C. $530 \, \text{g}\text{m}$
• D. $63 \, \text{g}\text{m}$

Answer 1

Answer: (A)

no. of equivalent of acid = no of equivalent of Base

$\left(\right. \frac{\text{W} \text{e} \text{i} \text{g} \text{h} \text{t}}{\text{M} \text{o} \text{l} \text{e} \text{c} \text{u} \text{l} \text{a} \text{r} \, \text{w} \text{e} \text{i} \text{g} \text{h} \text{t}} \times ' \text{n} ' \text{f} \text{a} \text{c} \text{t} \text{o} \text{r} \left.\right)_{\text{B} \text{a} \text{s} \text{e}} = \, \, \left(\right. \text{N} \times \left(\text{V}\right)_{\text{l} \text{i} \text{t} \text{e} \text{r}} \left.\right)_{\text{A} \text{c} \text{i} \text{d}}$

$\therefore $ n factor of NaOH = 1

$\therefore $ n factor of Oxalic acid = 1

$1 \, \, = \, \, \frac{\text{x} \, \, \times \, \, 1000}{63 \, \, \times \, \, 100} \, \text{x}= \, 6.3 \, \text{g}\text{m}\text{s}$