When $K M n O_{4}$ is titrated against ferrous ammonium sulphate in acid medium then the equivalent mass of $K M n O_{4}$ will be:

\frac{Molecular mass}{10}

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\frac{Molecular mass}{5}

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\frac{Molecular mass}{3}

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\frac{Molecular mass}{2}

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Solution

The correct option is A $\frac{Molecular mass}{5}$
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O n - f a c t o r = 5$ .
$E q u i v a l e n t m a s s = \frac{M o l e c u l a r m a s s}{n - f a c t o r} = \frac{M o l e c u l a r m a s s}{5}$

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$K M n O_{4}$ reacts with ferrous ammonium sulphate according to the equation $M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O$ , here 10 ml of 0.1 M $K M n O_{4}$ is equivalent to

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When KMnO4 is titrated against ferrous ammonium sulphate in acid medium then the equivalent mass of KMnO4 will be:

The correct option is A Molecular mass5MnO−4+8H++5e−→Mn2++4H2O n−factor=5.Equivalent mass=Molecular massn−factor=Molecular mass5

When $K M n O_{4}$ is titrated against ferrous ammonium sulphate in acid medium then the equivalent mass of $K M n O_{4}$ will be:

The correct option is A $\frac{Molecular mass}{5}$
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O n - f a c t o r = 5$ .
$E q u i v a l e n t m a s s = \frac{M o l e c u l a r m a s s}{n - f a c t o r} = \frac{M o l e c u l a r m a s s}{5}$