Equivalent weight of Baeyer's reagent is:

158

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31.6

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52.66

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Solution

The correct option is B $52.66$
i) Equivalent weight $= \frac{m o l e c u l a r w e i g h t}{c h a n g e i n o x i d a t i o n s t a t e}$
ii) Change in oxidation on state of ${K M n O}_{4}$ in basic medium $= 3$
iii) Molecular weight of ${K M n O}_{4} = 158$
Equivalent weight $= \frac{158}{3} = 52.66$ .

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