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Potassium Permanganate

If x moles ...

Question

If $x$ moles of $K I$ are oxidised by number of moles of ${K I O}_{3}$ formed when 1 mole of $I_{2}$ is boiled with excess of $K O H$ , then what is the value of $\frac{1}{x}$ ?
$6 K O H + 4 I_{2} ⟶ 5 K I + K I O_{3} + 3 H_{2} O$
$K I + K I O_{3} H^{+} - - \to I_{2} + H_{2} O$

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Solution

The correct option is D $3$
The reactions are given below:
$6 K O H + 4 I_{2} ⟶ 5 K I + K I O_{3} + 3 H_{2} O$
$K I + K I O_{3} H^{+} - - \to I_{2} + H_{2} O$
$3 I_{2} = {K I O}_{3}$
$1 I_{2} = \frac{1}{3} {K I O}_{3} = \frac{1}{3} K I$
Hence, $\frac{1}{3}$ is the value of $x$ .
Hence, the value of $\frac{1}{x}$ is $3$ .

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