Question

The equivalent weight of $K_{2}C{r}_2{O}_7$ (molecular weight = 294 u) in the following reaction is:
$K_{2}C{r}_2{O}_7+6KI+7H_{2}S{O}_4\to C{r}_2(S{O}_4{)}_3+3I_2+4K_{2}S{O}_4+7H_{2}O$

• A. $\frac{294}{6}$
• B. $\frac{294}{3}$
• C. $\frac{294}{4}$
• D. $\frac{294}{1}$

Answer 1

The correct option is A $\frac{294}{6}$

The reaction given here is,
$K_{2}C{r}_2{O}_7+6KI+7H_{2}S{O}_4\to C{r}_2(S{O}_4{)}_3+3I_2+4K_{2}S{O}_4+7H_{2}O$

Here, the oxidation state of $Cr$ is changing from +6 to +3.
Thus, the total moles of electrons gained by $Cr$ = $2\times 3=6$
Hence, n-factor = 6

$\text{Equivalent weight}=\frac{\text{molar mass}}{\text{n-factor}}$

$\Rightarrow \text{Equivalent weight}=\frac{294}{6}$