$3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O$
If molecular weight of $K C l O_{3}$ is $M$ , then equivalent weight of $K C l O_{3}$ is:

\frac{M}{4}

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\frac{M}{2}

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\frac{3 M}{2}

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\frac{3 M}{4}

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Solution

The correct option is C $\frac{3 M}{2}$
Here $C l O_{3}^{-}$ undergoes both oxidation and reduction to form $C l O_{4}^{-}$ and $C l O_{2}$ .
$C l O_{3}^{-} \to C l O_{4}^{-} + 2 e^{-} (x = 2) Oxidation e^{-} + C l O_{3}^{-} \to C l O_{2} (x = 1) reduction E_{w} = M + \frac{M}{2} = \frac{3 M}{2}$

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Similar questions

3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O

K C l O_{3}

M

K C l O_{3}

3 K C l O_{3} + 3 H_{2} S O_{4} \to 3 K H S O_{4} + H C l O_{4} + 2 C l O_{2} + H_{2} O

K C l O_{3}

3 M n O_{2} - ------- - + 6 K O H + K C l O_{3} ⟶ 3 K_{2} M n O_{4} + K C l + 3 H_{2} O

M

2 K M n O_{4} + 3 H_{2} S O_{4} ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 3 H_{2} O + 5 (O)

K M n O_{4}

K M n O_{4} = 158

K C l O_{3}

%

K C l O_{3}

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