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

$(\frac{\text{W}\text{e}\text{i}\text{g}\text{h}\text{t}}{\text{M}\text{o}\text{l}\text{e}\text{c}\text{u}\text{l}\text{a}\text{r}\text{w}\text{e}\text{i}\text{g}\text{h}\text{t}}{\times }^{\prime }{\text{n}}^{\prime }\text{f}\text{a}\text{c}\text{t}\text{o}\text{r}{)}_{\text{B}\text{a}\text{s}\text{e}}=(\text{N}\times {\left(\text{V}\right)}_{\text{l}\text{i}\text{t}\text{e}\text{r}}{)}_{\text{A}\text{c}\text{i}\text{d}}$

$\therefore$ n factor of NaOH = 1

$\therefore$ n factor of Oxalic acid = 1

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