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In the standardization of Na₂S₂O₃ using K₂Cr₂O₇ by iodometry, the equivalent weight of K₂Cr₂O₇ is

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(molecular weight)/3

(molecular weight)/6

same as molecular weight

(molecular weight)/2

answer is B.

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Detailed Solution

$K_{2} {\overset{+ 6}{Cr}}_{2} O_{7} \to 2 {Cr}^{+ 3}$

So change in oxidation state of Cr = +6

The equivalent weight of K₂Cr₂O₇ is $= \frac{M . Wt}{Valency}$ $= \frac{M}{6}$

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