Question

The equivalent weight of $K_{2}C{r}_2{O}_7$ as an oxidising agent in acidic medium is:

• A. $\frac{2}{3}$ of its molecular weight
• B. $\frac{1}{3}$ of its molecular weight
• C. $\frac{1}{6}$ of its molecular weight
• D. $\frac{1}{2}$of its molecular weight

Answer 1

The correct option is C $\frac{1}{6}$ of its molecular weight

${C{r}_2{O}_7}^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$

n-factor of $K_{2}C{r}_2{O}_7$ in acidic medium is $6$.

$\therefore Eq.wt.=\frac{Molecularweight}{6}$

Hence, option C is correct.