Equivalent conductance of an electrolyte cont | vedclass

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Question

(B) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution is the sum of the ionic conductance

Vedclass · Accepted Answer

$111.2$

Vedclass · Answer

(B) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution is the sum of the ionic conductances of the cation and the anion.For $NaF$,$\Lambda^{\infty}_{NaF} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{F^-} = 90.1 \, \Omega^{-1} \, cm^2$.For $KF$,$\Lambda^{\infty}_{KF} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{F^-}$.Since the ionic mobility of $K^+$ is higher than $Na^+$,the equivalent conductance of $KF$ will be higher than $NaF$.However,in many simplified textbook contexts,if the question implies a comparison where the values are expected to be similar or if it is a trick question regarding the nature of the electrolyte,we must note that $KF$ has a higher value than $90.1$.Given the options provided,$111.2$ is the only value greater than $90.1$ that fits the expected trend for the substitution of $Na^+$ with $K^+$.

Equivalent conductance of an electrolyte containing $NaF$ at infinite dilution is $90.1 \, \Omega^{-1} \, cm^2$. If $NaF$ is replaced by $KF$,what is the value of equivalent conductance? ........... $\Omega^{-1} \, cm^{2}$

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