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Solve the following.
Answer the following: Ionic conductance at infinite dilution of and ions are and respectively…
Answer the following:
(a)[1.5]
Ionic conductance at infinite dilution of $\text{Al}^{3+}$ and $\text{SO}_4^{2-}$ ions are $189\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}$ and $160\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}$ respectively.
Calculate the molar conductance of $\text{Al}_2(\text{SO}_4)_3$ at infinite dilution.
(b)[1.5]
The resistance of $0\cdot1\text{ N}$ solution of an electrolyte in a conducting cell was found to be $15\text{ ohm}$.
Calculate the following if the cell constant is $0\cdot367\text{ cm}^{-1}$.
(1) Conductance (C)
(2) Specific conductance ($\kappa$)
(3) Equivalent conductance ($\Lambda_{\text{eq}}$)
Answer
Answer (a)
AI$\Lambda^\circ_m[\mathrm{Al_2(SO_4)_3}] = 2\lambda^\circ(\mathrm{Al^{3+}}) + 3\lambda^\circ(\mathrm{SO_4^{2-}}) = 2(189) + 3(160) = 378 + 480 = 858$
Final answer: 858 ohm^-1 cm^2 mol^-1
Answer (b)
AIFormula used: Conductivity from resistance: kappa = Gstar/Rcell with Gstar = 0.367 cm^-1, Rcell = 15 ohm, so kappa = 0.02447 S cm^-1
(1) $C = \dfrac{1}{R} = \dfrac{1}{15} = 0.0667\ \mathrm{ohm^{-1}}$ (S)
(2) $\kappa = G^* \times C = 0.367 \times 0.0667 = 0.0245\ \mathrm{ohm^{-1}\ cm^{-1}}$
(3) $\Lambda_{eq} = \dfrac{\kappa \times 1000}{N} = \dfrac{0.02447 \times 1000}{0.1} = 244.7\ \mathrm{ohm^{-1}\ cm^2\ eq^{-1}}$
Final answer: 244.7 S cm^2 eq^-1
From ISC 2026 Improvement Chemistry Paper 1, question 18(ii).
Check your working with the Chemistry numericals calculator: solutions, electrochemistry and kinetics formulas: fill in what you know and get the rest.