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Answer the following.
Kohlrausch observed an interesting pattern between the values of molar conductance at infinite…
Kohlrausch observed an interesting pattern between the values of molar conductance at infinite dilution ($\Lambda^\circ_m$) for strong electrolytes. It was observed that different pairs of electrolytes having a common cation or anion had almost same difference of $\Lambda^\circ_m$. On the basis of his observation, he postulated a law known as Kohlrausch’s Law of Independent Migration of ions.
The values of molar conductivities at infinite dilution for some cations and anions are as follows:
| Ion | $\lambda^\circ_m (\text{S cm}^2\text{mol}^{-1})$ |
|---|---|
| $\text{Ba}^{2+}$ | 127.2 |
| $\text{Cl}^-$ | 76.3 |
| $\text{Ca}^{2+}$ | 119.0 |
| $\text{SO}_4^{2-}$ | 160.0 |
(i)[1.0]
State the Kohlrausch’s Law of Independent Migration of ions.
(ii)[1.0]
Calculate the molar conductance at infinite dilution $\Lambda^\circ_m$ for $\text{BaCl}_2$.
(iii)[1.0]
Arrange the values of $\Lambda^\circ_m$ for $\text{CaSO}_4$ and $\text{BaCl}_2$ in increasing order.
Answer
Answer (i)
AIKohlrausch's Law of Independent Migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the cation and the anion of the electrolyte:
$\Lambda^\infty_m = \nu_+ \lambda^\infty_+ + \nu_- \lambda^\infty_-$
where $\nu_+$ and $\nu_-$ are the number of cations and anions per formula unit of the electrolyte.
Answer (ii)
AIAccording to Kohlrausch's law:
$\Lambda^\infty_m(\text{BaCl}_2) = \lambda^\infty_m(\text{Ba}^{2+}) + 2 \lambda^\infty_m(\text{Cl}^-)$
Substituting the given values:
$\Lambda^\infty_m(\text{BaCl}_2) = 127\cdot2\text{ S cm}^2\text{mol}^{-1} + 2 \times 76\cdot3\text{ S cm}^2\text{mol}^{-1}$
$= 127\cdot2 + 152\cdot6 = 279\cdot8\text{ S cm}^2\text{mol}^{-1}$
Final answer: 279.8 $\text{S cm}^{2} \text{mol}^{-1}$
Answer (iii)
AI$\Lambda^\infty_m(\text{CaSO}_4) = \lambda^\infty_m(\text{Ca}^{2+}) + \lambda^\infty_m(\text{SO}_4^{2-}) = 119\cdot0 + 160\cdot0 = 279\cdot0\text{ S cm}^2\text{mol}^{-1}$
$\Lambda^\infty_m(\text{BaCl}_2) = 279\cdot8\text{ S cm}^2\text{mol}^{-1}$
Increasing order:
$\Lambda^\infty_m(\text{CaSO}_4) < \Lambda^\infty_m(\text{BaCl}_2)$
(or $279\cdot0\text{ S cm}^2\text{mol}^{-1} < 279\cdot8\text{ S cm}^2\text{mol}^{-1}$).
From ISC 2026 Chemistry Paper 1, question 1(C).
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