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Introduction to Oxidation and Reduction

KMn O 4 react...

Question

$K M n O_{4}$ reacts with $K I$ in basic medium to form $I_{2}$ and $M n O_{2}$ . When 250 mL of 0.1 M $K I$ solution is mixed with 250 mL of 0.02 M $K M n O_{4}$ in basic medium, what is the number of moles of $I_{2}$ formed?

A

0.015

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B

0.0075

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C

0.005

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D

0.01

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Solution

The correct option is B 0.0075
Number of milli-equivalents of $K M n O_{4}$ = Number of milli-equivalents of $I_{2}$ = $0.02 \times 3 \times 250 = 0.015$

Number of milli-moles of $I_{2}$ is $\frac{0.015}{2} = 0.0075$ .

So, the correct option is $B$

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