Question

In the standardisation of Na${}_2{S}_2{O}_3$ 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

Answer: (B)

The following reaction occur between $S_2{O}_3^{2-}$ and $C{r}_2{O}_7^{2-}:$
$26H^{+}+3S_2{O}_3^{2-}+4C{r}_2{O}_7^{2-}\to 6S{O}_4^{2-}+8C{r}^{3+}+13H_{2}O$
Change in oxidation number of $C{r}_2{O}_7^{2-}$ per formula unit is $6$ (it is always fixed for $C{r}_2{O}_7^{2-}$
Hence equivalent weight of $K_{2}C{r}_2{O}_7=\frac{\text{Molecular weight}}{6}$