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Standard XII

Chemistry

Types of Redox Reactions

In a volumetr...

Question

In a volumetric experiment, it was found that a solution of $K M n O_{4}$ is reduced to $M n S O_{4}$ . If the normality of solution is $1 N$ , then molarity of solution will be:

0.5 M

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0.2 M

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1 M

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0. 4 M

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Solution

The correct option is B $0.2 M$
$N o r m a l i t y = M o l a r i t y \times$ Change in oxidation number
$+ 7 K M n O_{4} \to + 2 M b S O_{4}$
Change in oxidation number $= 5$
$∴ N = M \times 5$
$1 = M \times 5$
$M = 0.2$ .

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Similar questions

Q. A solution of

K M n O_{4}

is reduced to

M n O_{2}

. The normality of solution is

0.6

. The molarity is:

Q. A sample of

M n S O_{4} \cdot 4 H_{2} O

is strongly heated in air. The residue

(M n_{3} O_{4})

left was dissolved in

100 m L

0.1 N F e S O_{4}

containing

d i l . H_{2} S O_{4}

. This solution was completely reacted with

50 m L

K M n O_{4}

solution.

25 m L

of this

K M n O_{4}

solution was completely reduced by

30 m L

0.1 N F e S O_{4}

solution. Calculate the mass of

M n S O_{4} \cdot 4 H_{2} O

in sample. (in g as nearest integer)

Q. Calculate the molarity of the resultant solution in each of the following cases:
i. 200 mL of 1 M HCl + 300 mL water
ii. 200 mL of 1 M HCl + 100 mL of 0.5 M HCl

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 $K M n O_{4} = 158 g m o l^{- 1}$ )

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In a volumetric experiment, it was found that a solution of KMnO4 is reduced to MnSO4. If the normality of solution is 1 N, then molarity of solution will be:

The correct option is B 0.2 MNormality=Molarity× Change in oxidation number+7KMnO4→+2MbSO4Change in oxidation number =5∴N=M×51=M×5M=0.2.

In a volumetric experiment, it was found that a solution of $K M n O_{4}$ is reduced to $M n S O_{4}$ . If the normality of solution is $1 N$ , then molarity of solution will be:

The correct option is B $0.2 M$
$N o r m a l i t y = M o l a r i t y \times$ Change in oxidation number
$+ 7 K M n O_{4} \to + 2 M b S O_{4}$
Change in oxidation number $= 5$
$∴ N = M \times 5$
$1 = M \times 5$
$M = 0.2$ .