Select the correct statement(s).

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$\land_{m}^{\infty}$ (molar conductance at infinite dilution) of strong electrolytes as well as weak electrolytes can be obtained by Kohlraush’s law $\land_{m}^{\infty} = x_{+} λ_{+}^{\infty} + y_{-} λ_{-}^{\infty}$ for $A_{x} B_{y}$

$\land_{A g C l}^{\infty}$ can be obtained by known value of $\land_{A g N O_{3}}^{\infty}, \land_{H C l}^{\infty} a n d \land_{H N O_{3}}^{\infty}$

$\land_{m (B a S O_{4})}^{\infty} = \land_{m (B a C l_{2})}^{\infty} + \land_{m (H_{2} S O_{4})}^{\infty} - 2 \land_{m (H C l)}^{\infty}$

$\land_{m}^{\infty}$ of strong as well as weak electrolytes can be obtained by extrapolation of graph between $\land_{m}$ and $\sqrt{c} t o \sqrt{c} \to 0$

answer is B, C, D.

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Detailed Solution

For strong electrolyte, variation in graph is uniform, hence $\land_{m}$ can be extrapolated to $\sqrt{C} \to 0$ , to obtain $\land_{m}^{\infty}$
In case of weak electrolyte, variation of $\land_{m}$ with $\sqrt{C}$ is rapid in lower concentration range (higher dilution) hence graph is not uniform and $\land_{m}^{\infty}$ cannot be obtained by extrapolation to $\sqrt{C} \to 0$ . Thus (a) is incorrect

${Al}_{2} (S O_{4}) = 2 A l^{3 +} + 3 S O_{4}^{2 -}$
$\land_{m}^{\infty} = 2 λ_{A l^{+}}^{\infty} + 3 λ_{S O_{4}^{2 -}}^{\infty}$

Thus (a) is true
C) Also $\land_{B a S O_{4}}^{\infty} = \land_{B a C l_{2}}^{\infty} + \land_{H_{2} S O_{4}}^{\infty} - 2 \land_{H C l}^{\infty}$
Thus, it is also correct based on Kohlrausch’s Law
d) $\land_{A g C l}^{\infty} = \land_{A g N O_{3}}^{\infty} + \land_{H C l}^{\infty} - \land_{H N O_{3}}^{\infty} = \land_{A g^{+}}^{\infty} + \land_{c l^{-}}^{\infty}$
Thus, correct based on Kohlrausch’s law