50 cc of oxalic acid is oxidized by 25 cc of 0.20 N KMnO4. The mass of oxalic acid present in 500 cc of the solution is

Question

50 cc of oxalic acid is oxidized by 25 cc of 0.20 N KMnO4. 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

Tardigrade Admin · Accepted Answer

N1V1=N2V2 N1 × 50 = 0.20 × 25 N1 = 0.1 Now, Normality = (w2/ textEq. wt. × textVolume in mL or cc ) Hence, mass of oxalic acid in 500 cc of solution can be calculated as 0.1 = (w2 × 1000/63 × 500) w2 = 3.15 g

Q. $50 cc$ of oxalic acid is oxidized by $25 cc$ of $0.20 N K M n O_{4}$ . The mass of oxalic acid present in $500 cc$ of the solution is

3.15 g

31.5 g

6.3 g

63 g

$N_{1} V_{1} = N_{2} V_{2}$
$N_{1} \times 50 = 0.20 \times 25$
$N_{1} = 0.1$
Now,
Normality = $\frac{w _{2}}{Eq. wt. \times Volume in mL or cc}$
Hence, mass of oxalic acid in $500 cc$ of solution can be calculated as
$0.1 = \frac{w _{2} \times 1000}{63 \times 500}$
$w_{2} = 3.15 g$

Q. 50cc of oxalic acid is oxidized by 25cc of 0.20NKMnO4​. The mass of oxalic acid present in 500cc of the solution is