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  4. The equivalent weight of K 2 Cr 2 O 7 in acidic medium

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Q. The equivalent weight of $K _{2} Cr _{2} O _{7}$ in acidic medium

The d-and f-Block Elements

Solution:

$K _{2} Cr _{2} O _{7}+3 H _{2} SO _{4} \rightarrow K _{2} SO _{4}+ Cr _{2}\left( SO _{4}\right)_{3}+3( O )+3 H _{2}$
No. of electrons lost $=12-6=6$
$\therefore$ Equivalent weight $=\frac{M}{6}=\frac{294}{6}=49 .$