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The equivalent conductance of \frac{M}{32} solution of a weak monobasic acid is 8.0 mho cm2 and at infinite dilution is 400 mho cm2. The dissociation constant of this acid is:

 

  • Option 1)

    1.25 x 10-6

  • Option 2)

    6.25 x 10-4

  • Option 3)

    1.25 x 10-4

  • Option 4)

    1.25 x 10-5

 

Answers (1)

best_answer

 

Application of Kohlrausch's law -

Determination of Degree of dissociation of weak electrolytes.

- wherein

Degree of dissociation

\alpha = \frac{\Lambda m}{\Lambda_{m}^{0} }

 

 \Lambda = 8 \ mho \ cm^{+2}

\Lambda_\infty = 400 \ mho \ cm^{2}

Degree of dissociation \alpha = \Lambda /\Lambda_\alpha - 8/400 

= 2 \times 10-2

Dissocation constant

K = C\alpha ^2 (C= M/32)

K= 1/32 \times 2 \times 10-2 2 \times 10-2 

K = 1.25 \times 10-5 


Option 1)

1.25 x 10-6

incorrect

Option 2)

6.25 x 10-4

incorrect

Option 3)

1.25 x 10-4

incorrect

Option 4)

1.25 x 10-5

correct

Posted by

divya.saini

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