Given below are two statements: Statement I: Mohr's salt is composed of only three types of ions-ferrous, ammonium and sulphate. Statement II: If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are mathbfx_1$\mathbf{x}_1$, mathbfx_2$\mathbf{x}_2$ and mathbfx_3$\mathbf{x}_3$ mathrmS\ cm^2\ mathrmmol^-1$\mathrm{S\ cm}^2\ \mathrm{mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by mathbfx_1 + mathbfx_2 + 2mathbfx_3$\mathbf{x}_1 + \mathbf{x}_2 + 2\mathbf{x}_3$. In the light of the given statements, choose the correct answer from the options given below:

### Related Formula lambda_m^infty = nu_+ lambda_+^infty + nu_- lambda_-^infty$$\lambda_m^{\infty} = \nu_+ \lambda_+^{\infty} + \nu_- \lambda_-^{\infty}$$ ### Core Logic Statement I: Mohr's salt is a double salt with chemical formula: mathrmFeSO_4 cdot (NH_4)_2SO_4 cdot 6H_2O$$\mathrm{FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O}$$ When dissolved in water, it completely dissociates into three distinct ionic species: mathrmFe^2+ text (ferrous), quad mathrmNH_4^+ text (ammonium), quad textand mathrmSO_4^2- text (sulphate)$$\mathrm{Fe}^{2+} \text{ (ferrous)}, \quad \mathrm{NH}_4^+ \text{ (ammonium)}, \quad \text{and } \mathrm{SO}_4^{2-} \text{ (sulphate)}$$ Thus, Statement I is true. Statement II: According to Kohlrausch's law of independent migration of ions: lambda_m^infty(textMohr's Salt) = 1 cdot lambda_m^infty(mathrmFe^2+) + 2 cdot lambda_m^infty(mathrmNH_4^+) + 2 cdot lambda_m^infty(mathrmSO_4^2-)$$\lambda_m^{\infty}(\text{Mohr's Salt}) = 1 \cdot \lambda_m^{\infty}(\mathrm{Fe}^{2+}) + 2 \cdot \lambda_m^{\infty}(\mathrm{NH}_4^+) + 2 \cdot \lambda_m^{\infty}(\mathrm{SO}_4^{2-})$$ lambda_m^infty = x_1 + 2x_2 + 2x_3$$\lambda_m^{\infty} = x_1 + 2x_2 + 2x_3$$ Statement II claims the expression is x_1 + x_2 + 2x_3$x_1 + x_2 + 2x_3$ (missing the coefficient 2$2$ for ammonium). Thus, Statement II is false. ### Pattern Recognition Kohlrausch's law matches stoichiometric coefficients directly to the ion quantities released. Mohr's salt formula contains (NH_4)_2$(NH_4)_2$, requiring a multiplier of 2$2$ for ammonium ion conductance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: d- and f-Block Elements