Question

What is the correct expression for the molar conductivity (\( \Lambda^\circ_m \)) of Al\(_2\)(SO\(_4\))\(_3\) at infinite dilution?

• A. \( 2\lambda^\circ(\text{Al}^{3+}) + 3\lambda^\circ(\text{SO}_4^{2-}) \)
• B. \( 3\lambda^\circ(\text{Al}^{3+}) + 2\lambda^\circ(\text{SO}_4^{2-}) \)
• C. \( \lambda^\circ(\text{Al}^{3+}) + \lambda^\circ(\text{SO}_4^{2-}) \)
• D. \( 6(\lambda^\circ(\text{Al}^{3+}) + \lambda^\circ(\text{SO}_4^{2-})) \)

Hint:

Always use the stoichiometric coefficients of dissociated ions while applying Kohlrausch’s Law to compute molar conductivity at infinite dilution.

Answer 1

The Correct Option is A

Step 1: Understand the dissociation of Al₂(SO₄)₃
Aluminium sulfate is a strong electrolyte and fully dissociates in water as: \[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-} \] So it produces 2 Al\(^{3+}\) ions and 3 SO₄^2- ions per formula unit. Step 2: Apply Kohlrausch’s Law of Independent Migration of Ions
According to Kohlrausch’s Law, the molar conductivity at infinite dilution is given by: \[ \Lambda^\circ_m = n_+\lambda^\circ_+ + n_-\lambda^\circ_- \] Where:
\( n_+ \) and \( n_- \) are the number of cations and anions respectively.
\( \lambda^\circ_+ \) and \( \lambda^\circ_- \) are the molar conductivities at infinite dilution of cation and anion. Step 3: Substitute for Al₂(SO₄)₃
\[ \Lambda^\circ_m = 2\lambda^\circ(\text{Al}^{3+}) + 3\lambda^\circ(\text{SO}_4^{2-}) \] Step 4: Conclusion
This matches option (1), hence it is the correct answer.