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

Question

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

Tardigrade Admin · Accepted Answer

MnO4-+8H++5e- arrow Mn2 ++4H2O Equivalent weight of KMnO4 in acidic medium =(m o l e c u l a r w e i g h t/5)=(158/5)=31.6

Q. $K M n O_{4}$ (mol. wt. = 158) oxidizes oxalic acid in acidic medium to $C O_{2}$ and water as follows

$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O$

What is the equivalent weight of $K M n O_{4}$ ?

158

31.6

39.5

79

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Equivalent weight of $K M n O_{4}$ in acidic medium $= \frac{m o l ec u l a r w e i g h t}{5} = \frac{158}{5} = 31.6$

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

158

31.6

39.5

79

MnO4−​+8H++5e−→Mn2++4H2​O Equivalent weight of KMnO4​ in acidic medium =5molecularweight​=5158​=31.6

Q. $K M n O_{4}$ (mol. wt. = 158) oxidizes oxalic acid in acidic medium to $C O_{2}$ and water as follows

$5 C_{2} O_{4}^{2 -} + 2 M n O_{4}^{-} + 16 H^{+} \to 10 C O_{2} + 2 M n^{2 +} + 8 H_{2} O$

What is the equivalent weight of $K M n O_{4}$ ?

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

Equivalent weight of $K M n O_{4}$ in acidic medium $= \frac{m o l ec u l a r w e i g h t}{5} = \frac{158}{5} = 31.6$