The mass of oxalic acid crystals required to prepare $50 mL$ of a $0.2 N$ solution is:

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Solution

Step 1:Finding gram equivalent weight of Oxalic acid
Oxalic acid exists as the dihydrate $H_{2} C_{2} O_{4} . 2 H_{2} O$
Molecular weight of oxalic acid dihydrate = $126 g {mol}^{- 1}$
$Gram equivalent weight = \frac{Molecular weight}{basicity}$
$Gram equivalent weight = \frac{126}{2}$
$Gram equivalent weight = 63 g {mol}^{- 1}$
Step 2: Finding mass of oxalic acid crystals required to prepare $50 mL$ of a $0.2 N$
$1$ Normal solution of oxalic acid should have $63 g$ in $1 L$ .
$0.2$ Normal solution of oxalic acid should have $= 60 \times 0.2 g = 12.6 g$
$0.2$ Normal solution of oxalic acid should have $12.6 g$ in $1 L$ .
$0.2$ Normal solution of oxalic acid should have in $50 mL = 0.05 L$ is calculated below.
= $12.6 \times 0.05 = 0.63 g$ in $50$ $mL$ .
Hence the mass of oxalic acid crystals required to prepare $50 mL$ of a $0.2 N$ solution is $0.63 g$

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