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)

3.15 g

Answer 2

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