Question

Find the equivalent weight of $K_{2}C{r}_2{O}_7^{2-}$ in the given reaction, if the molar mass of $C{r}_2{O}_7^{2-}$ is 'M' ?
$K_{2}C{r}_2{O}_7^{2-}\stackrel{H^{+}}{\to }C{r}^{3+}$

• A. $\frac{M}{6}$
• B. $\frac{M}{3}$
• C. $M$
• D. $2M$

Answer 1

The correct option is A $\frac{M}{6}$

In acidic medium, the balanced reaction will be:
$6e^{-}+\stackrel{+6}{C}{r}_2{O}_7^{2-}+14H^{+}\to 2\stackrel{+3}{C}{r}^{3+}+7H_{2}O$
n -factor $n_{f}of{K}_{2}C{r}_2{O}_7=|3-6|\times 2=6$
$\text{Equivalent weight of}{K}_{2}C{r}_2{O}_7=\frac{Mol.wtof{K}_{2}C{r}_2{O}_7}{6}=\frac{M}{6}$