$K_{2} {C r}_{2} O_{7} - -------- - + 7 H_{2} S O_{4} + 6 F e S O_{4} ⟶ K_{2} S O_{4} + {C r}_{2} {(S O_{4})}_{3} + 3 {F e}_{2} {(S O_{4})}_{3} + 7 H_{2} O$
The equivalent weight of the underlined species (by taking its molecular weight as $M$ ) is :

E = \frac{M}{6}

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E = \frac{M}{4}

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E = \frac{M}{3}

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none of these

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Solution

The correct option is A $E = \frac{M}{6}$
The oxidation number of Cr in $K_{2} C r_{2} O_{7}$ is $+ 6$ .
The oxidation number of Cr in $C r_{2} (S O_{4}) 3$ is $+ 3$ .
Thus, the oxidation number of Cr changes by $3$ .
Total change in the oxidation number of two Cr atoms is $6$ .
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{6}$

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