Question

In the standardization of ${Na}_2{S}_2{O}_3$ using $K_2{Cr}_2{O}_7$by iodometry, the equivalent weight of $K_2{Cr}_2{O}_7$ 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

since $K_2{Cr}_2{O}_7$ accepts 6 electrons for its reduction to $C{r}^{3+}$ ions $\therefore Eq.wt=Mol.wt$