Question

In the standardization of $N{a}_2{S}_2{O}_3$ 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. $Molecularweight$

Answer 1

Answer: (B)

$\frac{Molecularweight}{6}$

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

The reaction between $N{a}_2{S}_2{O}_3$ and $K_{2}C{r}_2{O}_7$ is as shown below.

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

The oxidation state of chromium in $K_{2}C{r}_2{O}_7$ changes from +6 to +3. The net change in oxidation number per formula unit is 6.

Hence, the value of n-factor is 6.

The equivalent weight $E=\frac{Molecularweight}{nfactor}=\frac{Molecularweight}{6}$.

Hence, the correct option is $B$