In an oxidation reduction reaction, $M n O_{4}^{-}$ ion is converted to $M n^{2 +}$ . What is the number of equivalents of $K M n O_{4}$ (mol. wt. $= 158$ ) present in $250 m L$ of $0.04 N$ $K M n O_{4}$ solution?

0.02

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0.05

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0.04

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0.07

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Solution

The correct option is A $0.05$
${M n O_{4}}^{-} + 8 H^{+} + 5 e^{⟶} M n^{2 +} + 4 H_{2} O$

Change in oxidation state is +7-+2=5(n)

Let the mass be m
Molarity is $\frac{n}{V}$ where
n =no. of moles= $\frac{m}{158}$
V=volume of solution in litre $= \frac{250}{1000} = 0.25$
Molarity= $0.04$
$⟹ m = 1.58 g$

Normality is $\frac{n_{e}}{V}$ where
$n_{e}$ is no of equivalents
V is volume of solution in liter

$N o r m a l i t y = m o l a r i t y \times n = 5 \times 0.04 = 0.2$
No of equivalents is $0.2 \times 0.25 = 0.05$

Hence, the correct option is $A$

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Similar questions

Q. In an oxidation-reduction reaction,

{M n O}_{4}^{-}

is converted to

{M n}^{+ 2}

. What is the number of equivalents of

{K M n O}_{4}

(mol. wt = 158) present in 250 ml of 0.04 M

{K M n O}_{4}

solution?

K M n O_{4}

(mol. Wt

= 158

) oxidizes acid in acidic medium to

C O_{2}

and water as follows.

5 C_{2} O_{4}^{2 -} + 2 M n O^{-} + 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}

What is the normality of a $K M n O_{4}$ solution in which $15.8 g$ of $K M n O_{4}$ is present in $250 m L$ solution and the solution is used as an oxidant, where ${M n O_{4}}^{-}$ is reduced to ${M n}^{2 +}$ ? (Given: molar mass of $K M n O_{4} = 158 g m o l^{- 1}$ )

Q. The following reaction occurs in acidic medium

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

What is the equivalent weight of

K M n O_{4}

? (Molecular weight of

K M n O_{4} = 158

)

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In an oxidation reduction reaction, MnO−4 ion is converted to Mn2+. What is the number of equivalents of KMnO4(mol. wt.=158) present in 250 mL of 0.04 N KMnO4 solution?

In an oxidation reduction reaction, $M n O_{4}^{-}$ ion is converted to $M n^{2 +}$ . What is the number of equivalents of $K M n O_{4}$ (mol. wt. $= 158$ ) present in $250 m L$ of $0.04 N$ $K M n O_{4}$ solution?