Question

Amount of oxalic acid required to prepare $250mL$ of $N/10$ solution (Mol. mass of oxalic acid $=126$) is:

• A. $1.5759g$
• B. $3.15g$
• C. $15.75g$
• D. $63.0g$

Answer 1

The correct option is A $1.5759g$

Normality $=\frac{amountofacid}{63}\times vol.\left(L\right)$

$0.1=\frac{amountacid}{63}\times \frac{1000}{250}[\because \frac{126}{2}=63]$

$AmountofO.A\Rightarrow \frac{0.1}{4}\times 63=1.5759g$