Question

$\lambda_{m}^{\circ}$ values for cation and anion of a binary electrolyte (AB-type) is 120 $\frac{S cm^2}{mol}$ and 210 $\frac{S cm^2}{mol}$. $\Lambda_{m}^{\circ}$ for the electrolyte is, (in $\frac{S cm^2}{mol}$)

Answer 1

330 $\frac{S cm^2}{mol}$

Answer 2

The problem asks for the limiting molar conductivity ($\Lambda_{m}^{\circ}$) of a binary electrolyte (AB-type) given the limiting molar conductivities of its cation and anion. This can be determined using Kohlrausch's Law of Independent Migration of Ions.

Kohlrausch's Law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions, each multiplied by the number of ions produced by one formula unit of the electrolyte.

For a binary electrolyte of type AB, which dissociates into one cation ($A^+$) and one anion ($B^-$):

$AB \rightleftharpoons A^+ + B^-$

The limiting molar conductivity of the electrolyte ($\Lambda_{m}^{\circ}$) is given by:

$\Lambda_{m}^{\circ}(AB) = \nu_+ \lambda_{m}^{\circ}(A^+) + \nu_- \lambda_{m}^{\circ}(B^-)$

where $\nu_+$ and $\nu_-$ are the number of cations and anions produced per formula unit, respectively. For an AB-type electrolyte, $\nu_+ = 1$ and $\nu_- = 1$.

So, the formula simplifies to:

$\Lambda_{m}^{\circ}(AB) = \lambda_{m}^{\circ}(cation) + \lambda_{m}^{\circ}(anion)$

Given values:

Limiting molar conductivity of cation ($\lambda_{m}^{\circ}(cation)$) = 120 S cm$^2$/mol

Limiting molar conductivity of anion ($\lambda_{m}^{\circ}(anion)$) = 210 S cm$^2$/mol

Substitute these values into the formula:

$\Lambda_{m}^{\circ}(AB) = 120 \, \frac{S \, cm^2}{mol} + 210 \, \frac{S \, cm^2}{mol}$

$\Lambda_{m}^{\circ}(AB) = 330 \, \frac{S \, cm^2}{mol}$