Question

$50\, cc$ of oxalic acid is oxidized by $25 \,cc$ of $0.20 \, N \, KMnO_4$. The mass of oxalic acid present in $500\, cc$ of the solution is

• A. 3.15 g
• B. 31.5 g
• C. 6.3 g
• D. 63 g

Answer 1

Answer: (A)

$N_1V_1=N_2V_2$
$N_1 \times 50 = 0.20 \times 25$
$N_1 = 0.1$
Now,
Normality = $\frac{w_2}{\text{Eq. wt.} \times \text{Volume in mL or cc }}$
Hence, mass of oxalic acid in $500\, cc$ of solution can be calculated as
$0.1 = \frac{w_2 \times 1000}{63 \times 500}$
$w_2 = 3.15 \, g$