Write the formula of Mohr’s salt

Mohr’s salt is $\mathrm{FeSO_4\cdot (NH_4)_2SO_4\cdot 6H_2O}$ or equivalently, $\mathrm{(NH_4)_2Fe(SO_4)_2\cdot 6H_2O}$

Check Statement I

In aqueous solution, Mohr’s salt dissociates as: $\mathrm{(NH_4)_2Fe(SO_4)_2 \rightarrow Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}}$

The water of crystallization does not count as ions.

So, the ions present are of only three types:

• $\mathrm{Fe^{2+}}$ (ferrous)
• $\mathrm{NH_4^+}$ (ammonium)
• $\mathrm{SO_4^{2-}}$ (sulfate)

Therefore, Statement I is true.

Check Statement II

Let the molar conductance at infinite dilution of ions be:

• $\lambda^\infty(\mathrm{Fe^{2+}})=x_1$
• $\lambda^\infty(\mathrm{NH_4^+})=x_2$
• $\lambda^\infty(\mathrm{SO_4^{2-}})=x_3$

By Kohlrausch’s law, the molar conductance at infinite dilution of Mohr’s salt is the sum of ionic conductances multiplied by their stoichiometric coefficients: $\Lambda_m^\infty = 1\cdot x_1 + 2\cdot x_2 + 2\cdot x_3$

because one formula unit gives: $\mathrm{Fe^{2+} + 2NH_4^+ + 2SO_4^{2-}}$

But Statement II says: $x_1 + x_2 + 2x_3$ which misses one $x_2$ term.

Therefore, Statement II is false.

Conclusion

• Statement I: True
• Statement II: False

Hence the correct option is: $\boxed{\text{B}}$