Question

In the standardisation of $N_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry, the equivalent weight of $K_{2}C{r}_2{O}_7$ is:

• A. $\frac{molecularweight}{2}$
• B. $\frac{molecularweight}{6}$
• C. $\frac{molecularweight}{3}$
• D. same as molecular weight

Answer 1

The correct option is B $\frac{molecularweight}{6}$

Balancing the Redox equation

$K_{2}C{r}_2{O}_7\to 2C{r}^{+3}$

+6 +3

so,change in oxidation state = 6

$\therefore$ equivalent weight $=\frac{\text{Molecular weight}}{6}$