Question

The molar conductivity of $NaCl,HCl$ and $C{H}_{3}COONa$ at infinite dilution are $126.45$, $426.16$ and $91Sc{m}^{2}mo{l}^{-1}$ respectively. The molar conductivity of $C{H}_{3}COOH$ at infinite dilution is:

• A. $201.28S$ $c{m}^{2}mo{l}^{-1}$
• B. $698.28S$ $c{m}^{2}mo{l}^{-1}$
• C. $390.71S$ $c{m}^{2}mo{l}^{-1}$
• D. $540.48S$ $c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is C $390.71S$ $c{m}^{2}mo{l}^{-1}$

The molar conductivity of $NaCl,HCl$ and $C{H}_{3}COONa$ at infinite dilution are $126.45,426.16$ and $91Sc{m}^{2}mo{l}^{-1}$ respectively. The molar conductivity of $C{H}_{3}COOH$ at infinite dilution is $390.71$ $S$$c{m}^{2}mo{l}^{-1}$.

The calculations are as shown below:

${\lambda }_{C{H}_{3}COOH}^{\infty }={\lambda }_{C{H}_{3}COONa}^{\infty }+{\lambda }_{HCl}^{\infty }-{\lambda }_{NaCl}^{\infty }$

${\lambda }_{C{H}_{3}COOH}^{\infty }=91Sc{m}^{2}mo{l}^{-1}+426.16Sc{m}^{2}mo{l}^{-1}-126.45Sc{m}^{2}mo{l}^{-1}$

${\lambda }_{C{H}_{3}COOH}^{\infty }=390.71Sc{m}^{2}mo{l}^{-1}$