Question

The molar conductivity of\[NaOH\], \[HCL\], and $C{{H}_{3}}COONa$ at infinite dilution are 126.45, 426.16 and 91 $Sc{{m}^{2}}mo{{l}^{-1}}$respectively. The molar conductivity of $C{{H}_{3}}COOH$at infinite dilution is:
A. $201.28Sc{{m}^{2}}mo{{l}^{-1}}$
B. $698.285Sc{{m}^{2}}mo{{l}^{-1}}$
C. $390.71Sc{{m}^{2}}mo{{l}^{-1}}$
D. $540.48Sc{{m}^{2}}mo{{l}^{-1}}$

Answer 1

Hint: To solve this question, you will use Kohlrausch law of independent migration of ions. According to this law, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar conductivities of ions constituting it.

Complete step by step answer:
According to Kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivities of its ions. As you may already know molar conductivity is denoted by $\Lambda $and limiting molar conductivity by ${{\Lambda }^{{}^\circ }}m$
Another important point you must remember is that the limiting molar conductivity is the molar conductivities of a solution at infinite dilution, that is when the concentration of the electrolyte approaches zero.
Hence,
${{\Lambda }^{{}^\circ }}(Nacl)=126.45Sc{{m}^{2}}mo{{l}^{-1}}$
$\Lambda {}^\circ (HCl)=426.16Sc{{m}^{2}}mo{{l}^{-1}}$
${{\Lambda }^{{}^\circ }}(C{{H}_{3}}COONa)=91Sc{{m}^{2}}mo{{l}^{-1}}$
So, for ${{\Lambda }^{{}^\circ }}(C{{H}_{3}}COOH)$you need to equate the given values as:
${{\Lambda }^{{}^\circ }}(C{{H}_{3}}COONa)+{{\Lambda }^{{}^\circ }}(HCl)-{{\Lambda }^{{}^\circ }}(NaCl)$
 =91 + 426.16 - 126.45
=517.16 - 126.45
${{\Lambda }^{{}^\circ }}(C{{H}_{3}}COOH)=390.71Sc{{m}^{2}}mo{{l}^{-1}}$

Thus, the correct option is (C).

Note:
To understand which limiting molar conductivities you need to add or subtract, you need to look at the individual ions. $C{{H}_{3}}COOH$ is comprised of $C{{H}_{3}}CO{{O}^{-}}$and${{H}^{+}}$. Thus, you need to keep the $C{{H}_{3}}CO{{O}^{-}}$ part of $C{{H}_{3}}COONa$ but remove the $N{{a}^{+}}$part. This can be done by subtracting $NaCl$. If you subtract $NaCl$, $C{{l}^{-}}$also gets subtracted from HCL and leaves behind the ${{H}^{+}}$part you need to form $C{{H}_{3}}COOH$. This way students can calculate the limiting molar conductivity of any compound if the limiting molar conductivities of the individual ions making it up can be obtained either directly or indirectly.