Question

Equivalent conductance of $NaCl, HCl$ and $C_2H_5COONa$ at infinete dilution are $126.45, 426.16$ and $91 \Omega ^{-1} cm^2$ , respectively. The equivalent conductance of $C_2H_5COOH$ is

• A. 201.28 $\Omega^{-1} \, cm^2$
• B. 390.71 $\Omega^{-1} \, cm^2$
• C. 698.28 $\Omega^{-1} \, cm^2$
• D. 540.48 $\Omega^{-1} \, cm^2$

Answer 1

Answer: (B)

390.71 $\Omega^{-1} \, cm^2$

Answer 2

By Kohlrausch's law
\hspace5mm λ_{∞} for \, NaCl \, = λ_{Na^+} \, + λ_{Cl^-} \, \, \, \, \, \, \, \, \, \, \, \, \, ...(i)
\hspace5mm λ_{∞} for \, HCl= λ_{H^+}+ λ_{Cl^-}\, \, \, \, \, \, \, \, \, \, \, ...(ii)
$λ_{∞} for \, C_2H_5COONa=λ_{Na^+}+ λ_{C_2H_5COO^-} \, \, \, \, \, \, \, ...(iii)$
So, $λ_{∞} for \, C_2H_5COOH$
On adding Eqs. (ii) and (iii) and then subtracting
E (i)
$=λ_{∞} of \, C_2H_5COONa+ λ_{∞} \, of \, HCl$
\hspace15mm -λ_{∞} \, for \, NaCl
\hspace15mm =(91+426.16-126.45) \Omega ^{-1} cm^2
\hspace15mm =390.71 \Omega^{-1} cm^2