Question

In the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry, the equivalent weight of $K_{2}C{r}_2{O}_7$ is

• A. (Molecular weight)/2
• B. (Molecular Weight)/6
• C. (Molecular weight)/3
• D. Same as molecular weight

Answer 1

The correct option is B (Molecular Weight)/6

$K_{2}C{r}_2{O}_7$ acts as an oxidising agent in the presence of dil.$H_{2}S{O}_4$
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
$Eq.W=\frac{MolecularWeight}{Numberof{e}^{-}gain}=\frac{M}{6}$