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The molar conductivities \Lambda ^{\circ}{_{NaOAc}}\: and \: \Lambda ^{\circ}{_{HCl}}  at infinite dilution in water at 25^{\circ}C are 91.0 and 426.2 S cm2/mol  respectively. To calculate  \Lambda ^{\circ}{_{HOAc}}, the additional value required is

  • Option 1)

    \Lambda ^{\circ}{_{H_{2}O}}

  • Option 2)

    \Lambda ^{\circ}{_{KCl}}

  • Option 3)

    \Lambda ^{\circ}{_{NaOH}}

  • Option 4)

    \Lambda ^{\circ}{_{NaCl}}


Answers (1)

As we learnt in

Kohlrausch law of independent migration of ions -

The law states that limiting molar conductivity of an electrolyte can be represented at the sum of the individual contributions of the anion and cation of the electrolyte.




CH_{3}COONa+HCl\rightarrow CH_{3}COOH+NaCl

From the reaction,

\Lambda ^{\circ}_{CH_{3}COONa}+\Lambda ^{\circ}_{HCl}=\Lambda ^{\circ}_{CH_{3}COOH}+\Lambda ^{\circ}_{NaCl}

or\; \; \Lambda ^{\circ}_{CH_{3}COOH}=\Lambda ^{\circ}_{CH_{3}COONa}+\Lambda ^{\circ}_{HCl}-\Lambda ^{\circ}_{NaCl}

Thus to calculate the value of \; \; \Lambda ^{\circ}_{CH_{3}COOH} one should know the value of  \; \; \Lambda ^{\circ}_{NaCl} along with \; \; \Lambda ^{\circ}_{CH_{3}COONa} and \; \; \Lambda ^{\circ}_{HCl}

Option 1)

\Lambda ^{\circ}{_{H_{2}O}}

This option is incorrect.

Option 2)

\Lambda ^{\circ}{_{KCl}}

This option is incorrect.

Option 3)

\Lambda ^{\circ}{_{NaOH}}

This option is incorrect.

Option 4)

\Lambda ^{\circ}{_{NaCl}}

This option is correct.

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