Chemistry - Mole Concept Question with Solution | TestHub

$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+6\mathrm{F}{\mathrm{e}}^{2+}(\mathrm{n}=1)$(Mohr's salt)$+14{\mathrm{H}}^{+}\to 2\mathrm{C}{\mathrm{r}}^{3+}+6\mathrm{F}{\mathrm{e}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

Equivalent of$\mathrm{F}{\mathrm{e}}^{2+}=$moles of Mohr's salt

$=$equivalent of${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$

$=500\times 10^{-3}\times 6\times 1=3$

Hence mole percent of Mohr's salt

$=\frac{3}{4}\times 100=75$