The volume of 0.1 M oxalic acid that can be completely oxidized by 20 ml of 0.025 M KMnO4 solution is

Question

The volume of 0.1 M oxalic acid that can be completely oxidized by 20 ml of 0.025 M KMnO4 solution is (A) 125 ml (B) 25 ml (C) 12.5 ml (D) 37.5 ml

Tardigrade Admin · Accepted Answer

N=( text molecular weight / text equivalent weight ) × M N KMnO 4=(158/31.6) × 0.025 =0.125 N O.A. =(126/63) × 0.1=0.2 ∵ N1 V1=N2 V2 ∴ 0.2 × V =0.125 × 20 V =(0.125 × 20/0.2) =12.5 mL

Q. The volume of $0.1 M$ oxalic acid that can be completely oxidized by $20 m l$ of $0.025 M K M n O_{4}$ solution is

$125 m l$

$25 m l$

$12.5 m l$

$37.5 m l$

$N = \frac{molecular weight}{equivalent weight} \times M$

$N_{K M n O_{4}} = \frac{158}{31.6} \times 0.025$

$= 0.125$

$N_{O . A .} = \frac{126}{63} \times 0.1 = 0.2$

$∵ N_{1} V_{1} = N_{2} V_{2}$

$∴ 0.2 \times V = 0.125 \times 20$

$V = \frac{0.125 \times 20}{0.2}$

$= 12.5 m L$

Q. The volume of 0.1M oxalic acid that can be completely oxidized by 20ml of 0.025MKMnO4​ solution is

125ml

25ml

12.5ml

37.5ml