The molar conductivities of HCl,NaCl,CH_{3}CO | vedclass

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Question

(A) Molar conductivity at infinite dilution $(\Lambda_{m}^{\circ})$ depends on the mobility of ions in the solution.$HCl$ is a strong acid and dissoci

Vedclass · Accepted Answer

$HCl > NaCl > CH_{3}COONa > CH_{3}COOH$

Vedclass · Answer

(A) Molar conductivity at infinite dilution $(\Lambda_{m}^{\circ})$ depends on the mobility of ions in the solution.$HCl$ is a strong acid and dissociates completely into $H^{ }$ and $Cl^{-}$ ions. $H^{ }$ ions have the highest ionic mobility.$NaCl$ is a strong electrolyte,dissociating into $Na^{ }$ and $Cl^{-}$ ions.$CH_{3}COONa$ is a strong electrolyte,dissociating into $CH_{3}COO^{-}$ and $Na^{ }$ ions.$CH_{3}COOH$ is a weak acid and dissociates only partially,resulting in fewer ions in the solution compared to strong electrolytes.Since $HCl$ provides the highly mobile $H^{ }$ ion,it has the highest molar conductivity.$NaCl$ has higher conductivity than $CH_{3}COONa$ because the $Cl^{-}$ ion has higher mobility than the $CH_{3}COO^{-}$ ion.$CH_{3}COOH$ has the lowest molar conductivity due to its weak dissociation.Thus,the correct order is $HCl > NaCl > CH_{3}COONa > CH_{3}COOH$.

The molar conductivities of $HCl$,$NaCl$,$CH_{3}COOH$,and $CH_{3}COONa$ at infinite dilution follow the order:

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