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Q.

In acidic medium, equivalent weight of K₂Cr₂O₇ (mol. wt.= M) is

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a

M/3

b

M/4

c

M/6

d

M/2

answer is C.

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Detailed Solution

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O$
Equivalent weight of $K_{2} {Cr}_{2} O_{7}$
$= \frac{Molecular Mass}{6} = \frac{294.2}{6} = \frac{M}{6}$

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