Question

At ${25}^{o}C$, the equivalent conductance at infinite dilution of $HCl,{CH}_{3}COONa$ and $NaCl$ are $426.1,91.0$ and $126.45{cm}^2{\Omega }^{-1}{eq}^{-1}$ respectively, the ${\wedge }_{\infty }$ for $C{H}_{3}COOH\left(in{cm}^{-1}{\Omega }^{-1}{eq}^{-1}\right)$ is:

• A. $391.6$
• B. $390.6$
• C. $389.6$
• D. $388.6$

Answer 1

The correct option is B $390.6$

Given that equivalent conductance at infinite dilution of $HCl,C{H}_{3}COONa$ and $NaCl$ are $426.1,91.0$ and $126.45{cm}^2{\Omega }^{-1}{eq}^{-1}$ respectively.

$\therefore {{\Lambda }_{\infty }}_{\left(HCl\right)}={{\lambda }^0}_{H^{+}}+{{\lambda }^0}_{{Cl}^{-}}=426.1{cm}^2{\Omega }^{-1}{eq}^{-1}$

${{\Lambda }_{\infty }}_{\left(C{H}_{3}COONa\right)}={{\lambda }^0}_{{Na}^{+}}+{{\lambda }^0}_{{C{H}_{3}COO}^{-}}=91{cm}^2{\Omega }^{-1}{eq}^{-1}$

${{\Lambda }_{\infty }}_{\left(NaCl\right)}={{\lambda }^0}_{{Na}^{+}}+{{\lambda }^0}_{{Cl}^{-}}=126.45{cm}^2{\Omega }^{-1}{eq}^{-1}$

For ${{\Lambda }_{\infty }}_{\left(C{H}_{3}COOH\right)}$,

${{\Lambda }_{\infty }}_{\left(C{H}_{3}COOH\right)}={{\lambda }^0}_{{C{H}_{3}COO}^{-}}+{{\lambda }^0}_{H^{+}}$

$\Rightarrow {{\Lambda }_{\infty }}_{\left(C{H}_{3}COOH\right)}=({{\lambda }^0}_{{C{H}_{3}COO}^{-}}+{{\lambda }^0}_{{Na}^{+}})-({{\lambda }^0}_{{Na}^{+}}+{{\lambda }^0}_{{Cl}^{-}})+({{\lambda }^0}_{H^{+}}+{{\lambda }^0}_{{Cl}^{-}})$

$\Rightarrow {{\Lambda }_{\infty }}_{\left(C{H}_{3}COOH\right)}=91-126.45+426.1=390.55{cm}^2{\Omega }^{-1}{eq}^{-1}\approx 390.6{cm}^2{\Omega }^{-1}{eq}^{-1}$

Hence, the correct option is B.