Question

In the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ 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. $Samaasmolecularweight$

Answer 1

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

We know that:

The expression for the equivalent weight is E = $\frac{Molecularweight}{nfactor}$

The reaction of standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ iodometry is as follows:

$26H^{+}+3S_2{O}_3^{2-}+4C{r}_2{O}_7^{2-}\to 6S{O}_4^{2-}+8C{r}^{3+}+13H_{2}O$

In this reaction, the oxidation state of chromium in $K_{2}C{r}_2{O}_7$ changes from +6 to +3.

Thus, the net change in oxidation number per formula unit is 6.

So, the $nfactor$ is 6.

So equivalent weight can be written as:

E = $\frac{Molecularweight}{nfactor}$ $\Rightarrow$ $\frac{Molecularweight}{6}$

Hence,correct option is (B).