Question

The molar conductivities of $HCl,NaCl,C{H}_{3}COOHandC{H}_{3}COONa$ at infinite dilution follow the order

• A. >>>
• B. >>>
• C. >>>
• D. >>>

Answer 1

The correct option is A >>>

The molar conductivities of the following ions at infinite dilution are as follows:
${\lambda }_{\infty }C{H}_{3}CO{O}^{-}=40.9\times {10}^{-4}S{m}^{2}mo{l}^{-1}{\lambda }_{\infty }N{a}^{+}=50.10\times {10}^{-4}S{m}^{2}mo{l}^{-1}{\lambda }_{\infty }C{l}^{-}=76.35\times {10}^{-4}S{m}^{2}mo{l}^{-1}{\lambda }_{\infty }{H}^{-}=349\times {10}^{-4}S{m}^{2}mo{l}^{-1}{\lambda }_{\infty }HCl={\lambda }_{H^{+}}+{\lambda }_{C{l}^{-}}{\lambda }_{C{H}_{3}COOH}={\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{H^{+}}{\lambda }_{NaCl}={\lambda }_{Na}+{\lambda }_{C{l}^{-}}{\lambda }_{C{H}_{3}COONa}={\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{N{a}^{+}}So,{\lambda }_{HCl}=425.35S{m}^2/mol{\lambda }_{C{H}_{3}COOH}=389.9S{m}^2/mol{\lambda }_{NaCl}=126.45S{m}^2/mol{\lambda }_{C{H}_{3}COONa}=91S{m}^2/molHence,\text{the order followed is}:{\lambda }_{HCl}>{\lambda }_{C{H}_{3}COOH}>{\lambda }_{NaCl}>{\lambda }_{C{H}_{3}COONa}$