Question

In acidic medium, the equivalent weight of $K_{2}C{r}_2{O}_7$ (Mol. wt. $=M$ ) is

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

Answer 1

Answer: (B)

Key Idea Equivalent weight

$=\frac{\text{ Molecular mass }}{\text{ Change in oxidation number per atom }}$

Potassium dichromate, $K_{2}C{r}_2{O}_7$ is a powerful oxidising agent. In acidic medium, its action can be represented as

$K_2\stackrel{+6}{C{r}_2}{O}_7+4H_{2}S{O}_4⟶K_{2}S{O}_4+\stackrel{+3}{C{r}_2}{\left(S{O}_4\right)}_3+4H_{2}O+3[O]$

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

Eq. wt. of $K_{2}C{r}_2{O}_7$

$=\frac{\text{ molecular weight of }{K}_{2}C{r}_2{O}_7}{2\times \text{ change in oxidation number }}$

$=\frac{M}{2\times 3}[\because$ Two Cr atoms are involved. $]$

$\therefore$ Eq. wt. of $K_{2}C{r}_2{O}_7=\frac{M}{6}$