Equivalent conductivity at infinite dilution | vedclass

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Question

(A) The molar conductivity at infinite dilution for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is the sum of the molar conductivities of its constituent

Vedclass · Accepted Answer

$135.9 \ S \ cm^{2} \ eq^{-1}$

Vedclass · Answer

(A) The molar conductivity at infinite dilution for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is the sum of the molar conductivities of its constituent ions: $\lambda_{m}^{\infty} = \lambda_{m}^{\infty} (oxalate^{2-}) + \lambda_{m}^{\infty} (Na^{+}) + \lambda_{m}^{\infty} (K^{+})$.Substituting the given values: $\lambda_{m}^{\infty} = 148.2 + 73.5 + 50.1 = 271.8 \ S \ cm^{2} \ mol^{-1}$.The $n$-factor for the salt $[(COO^{-})_{2} Na^{+} K^{+}]$ is $2$ because the oxalate ion has a charge of $-2$.Therefore, the equivalent conductivity at infinite dilution is $\lambda_{eq}^{\infty} = \frac{\lambda_{m}^{\infty}}{n-factor} = \frac{271.8}{2} = 135.9 \ S \ cm^{2} \ eq^{-1}$.

Equivalent conductivity at infinite dilution for sodium potassium oxalate $[(COO^{-})_{2} Na^{+} K^{+}]$ will be (given molar conductivities of oxalate, $K^{+}$ and $Na^{+}$ ions at infinite dilution are $148.2$, $50.1$, and $73.5 \ S \ cm^{2} \ mol^{-1}$ respectively).

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