Question

${\left(N{H}_4\right)}_{2}C{r}_2{O}_7⟶N_2+C{r}_2{O}_3+2H_{2}O$
$E_{C{r}_2{O}_3}=\frac{M}{x}$. The value of $x$ is:

Answer 1

${\left(N{H}_4\right)}_{2}C{r}_2{O}_7⟶N_2+C{r}_2{O}_3+2H_{2}O$
-3 +6 0 +3
As we know,
Equivalent weight $=\frac{molarweight}{n-factor}$
$E_{C{r}_2{O}_3}=\frac{M}{6}$
So, the value of $x$ is $6$.