Question

In standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry the equiavlent 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 $/6$

In the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$,reduction of $K_{2}C{r}_2{O}_7$ takes place in such a manner:${Cr}_2{O}_7^{-2}{+14H}^{+}+6e^{-}\to {2Cr}^{+3}+7H_{2}O$

6 electrons are involved in the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ so the equivalent weight of $K_{2}C{r}_2{O}_7$ is molecular weight/6.

Note: Equivalent weight = Molecular weight/(number of electrons lost or gained)