Given the limiting molar conductivity as $\Lambda _{\rm{m}}^0\left( {{\rm{HCl}}} \right) = 425.9{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} {\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\rm{mo}}{{\rm{l}}^{ - 1}}$ $\Lambda _{\rm{m}}^0\left( {{\rm{NaCl}}} \right) = 126.4{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} {\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\rm{mo}}{{\rm{l}}^{ - 1}}$ $\Lambda _{\rm{m}}^0\left( {{\rm{C}}{{\rm{H}}_3}{\rm{COONa}}} \right) = 91{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} {\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\rm{mo}}{{\rm{l}}^{ - 1}}$ The molar conductivity , at infinite dilution, of acetic acid (${\rm{in}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} {\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\rm{mo}}{{\rm{l}}^{ - 1}}$) will be

JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

1 of 1

Application +4 | -1

Given the limiting molar conductivity as

\Lambda _\mathrm{m}^0\left( {\mathrm{HCl}} \right) = 425.9{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

\Lambda _\mathrm{m}^0\left( {\mathrm{NaCl}} \right) = 126.4{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

\Lambda _\mathrm{m}^0\left( {\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}} \right) = 91{\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

The molar conductivity , at infinite dilution, of acetic acid (

\mathrm{in}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\Omega ^{ - 1}}{\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}

) will be

View answer and explanation

Correct option: B

390.5

Option B (390.5) is correct because it is calculated by applying Kohlrausch's law to find the molar conductivity of acetic acid at infinite dilution.