In order to prepare one litre one normal solution of KMnO4, how many grams of KMnO4 are required, if the solution to be used in acidic medium for oxidation?

Question

In order to prepare one litre one normal solution of KMnO4, how many grams of KMnO4 are required, if the solution to be used in acidic medium for oxidation? (A) 158 g (B) 31.60 g (C) 62.0 g (D) 790 g

Tardigrade Admin · Accepted Answer

In acidic medium, underset text(From KMnO4)MnO4- + 8H+ + 5e-arrow Mn2+ + 4H2O Thus, eq. wt. of KMnO4 in acidic medium = ( textMol. wt./5) = (158/5) = 31.6 ∴ 1 N solution can be prepared by dissolving 31.6 g KMnO4 in 1 L solution.

Q. In order to prepare one litre one normal solution of $K M n O_{4}$ , how many grams of $K M n O_{4}$ are required, if the solution to be used in acidic medium for oxidation?

$158 g$

$31.60 g$

$62.0 g$

$790 g$

In acidic medium,
$(From K M n O_{4}) M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Thus, eq. wt. of $K M n O_{4}$ in acidic medium $= \frac{Mol. wt.}{5}$
$= \frac{158}{5} = 31.6$
$∴ 1 N$ solution can be prepared by dissolving $31.6 g$ $K M n O_{4}$ in $1 L$ solution.

Q. In order to prepare one litre one normal solution of KMnO4​, how many grams of KMnO4​ are required, if the solution to be used in acidic medium for oxidation?

158g

31.60g

62.0g

790g