KMnO4 (mol. wt = 158) oxidizes oxalic acid in acidic medium to CO2and water as follows. 5C2O42- + 2MnO4- + 16H+ → 10CO2 + 2Mn2+ + 8H2O What is the equivalent weight of KMnO4?

Question

KMnO4 (mol. wt = 158) oxidizes oxalic acid in acidic medium to CO2and water as follows. 5C2O42- + 2MnO4- + 16H+ → 10CO2 + 2Mn2+ + 8H2O What is the equivalent weight of KMnO4? (A) 158 (B) 31.6 (C) 39.5 (D) 79

Tardigrade Admin · Accepted Answer

MnO 4-+8 H ++5 e - longrightarrow Mn 2+ +4 H 2 O Eq. wt of KMnO 4 in acidic medium =( textMol . wt /5) =(158/5)=31.6

Q. $K M n O_{4}$ (mol. wt $= 158$ ) oxidizes oxalic acid in acidic medium to $C O_{2}$
$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O$
and water as follows. What is the equivalent weight of $K M n O_{4}$ ?

158

31.6

39.5

79

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
Eq. wt of $K M n O_{4}$ in acidic medium $= \frac{M o l . wt}{5}$
$= \frac{158}{5} = 31.6$

Q. KMnO4​ (mol. wt =158) oxidizes oxalic acid in acidic medium to CO2​ 5C2​O42−​+2MnO4−​+16H+→10CO2​+2Mn2++8H2​O and water as follows. What is the equivalent weight of KMnO4​?

158

31.6

39.5

79

MnO4−​+8H++5e−⟶Mn2++4H2​O Eq. wt of KMnO4​ in acidic medium =5Mol.wt​ =5158​=31.6

Q. $K M n O_{4}$ (mol. wt $= 158$ ) oxidizes oxalic acid in acidic medium to $C O_{2}$
$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O$
and water as follows. What is the equivalent weight of $K M n O_{4}$ ?

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ M n^{2 +} + 4 H_{2} O$
Eq. wt of $K M n O_{4}$ in acidic medium $= \frac{M o l . wt}{5}$
$= \frac{158}{5} = 31.6$