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Molar conductivities (\Lambda^{\circ}m) at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 s cm2 mol-1 respectively.

\Lambdam for CH3COOH will be :

  • Option 1)

    180.5 S cm2 mol-1

  • Option 2)

    290.8 S cm2 mol -1

  • Option 3)

    390.5 S cm2 mol-1

  • Option 4)

    425.5 S cm2 mol-1

 

Answers (1)

best_answer

As we discussed in

\Lambda ^{0}_{NaCl} = 126.4\:S\:cm^{2}\:mol^{-1}

\Lambda ^{0}_{HCl} = 425.9\:S\:cm^{2}\:mol^{-1}

\Lambda ^{0}_{CH_{3}COONa} = 91.0\:S\:cm^{2}\:mol^{-1}

\Lambda ^{0}_{CH_{3}COOH} = \Lambda ^{0}_{CH_{3}COONa} + \Lambda^{0}_{HCl}-\Lambda^{0}_{NaCl}

                     = 91+425.9+126.4

                     = 390.5 \:S cm^{2} \:mol^{-1}


Option 1)

180.5 S cm2 mol-1

This option is incorrect.

Option 2)

290.8 S cm2 mol -1

This option is incorrect.

Option 3)

390.5 S cm2 mol-1

This option is correct.

Option 4)

425.5 S cm2 mol-1

This option is incorrect.

Posted by

Aadil

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