The limiting molar conductivities of NaI,NaNO | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar

Vedclass · Accepted Answer

$14.0$

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(B) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(Ag^+) + \lambda_{m}^{\infty}(I^-)$We can express this using the given values:$\lambda_{m}^{\infty}(AgI) = \lambda_{m}^{\infty}(AgNO_3) + \lambda_{m}^{\infty}(NaI) - \lambda_{m}^{\infty}(NaNO_3)$Substituting the given values:$\lambda_{m}^{\infty}(AgI) = 13.3 + 12.7 - 12.0$$\lambda_{m}^{\infty}(AgI) = 26.0 - 12.0$$\lambda_{m}^{\infty}(AgI) = 14.0 \, mS \, m^2 \, mol^{-1}$

The limiting molar conductivities of $NaI$,$NaNO_3$ and $AgNO_3$ are $12.7$,$12.0$ and $13.3 \, mS \, m^2 \, mol^{-1}$,respectively (all at $25^{\circ} C$). The limiting molar conductivity of $AgI$ at this temperature is $.... \, mS \, m^2 \, mol^{-1}$.

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