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are 43.0 and 65 $oh{m^{ - 1}}$ respectively, calculate the degree of dissociation of NaCl solution.

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Given, Normality of solution of $NaCl = \dfrac{1}{{10}}N$

Specific conductivity $\left( K \right) = 0.0092$

Ionic conductances of $N{a^ + }$ and $C{l^ - }$ are 43.0 and 65 $oh{m^{ - 1}}$.

To find the equivalent conductance in terms of concentration\[\left( {\Lambda _{eq}^c} \right)\]

Equivalent conductance is the conducting power of all ions furnished by an equivalent of an electrolyte in any solution.

Equivalent conductance \[\left( {\Lambda _{eq}^c} \right) = \dfrac{{Conductivity\left( K \right) \times 1000}}{{Normality{\text{ of solution}}}}\]

= $\dfrac{{0.0092 \times 1000}}{{0.1}}$

= 92 $oh{m^{ - 1}}$.

To find equivalent conductance at infinite dilution.

Equivalent conductance at infinite dilution \[\left( {\Lambda _{eq}^\infty } \right)\] for $NaCl$ $ = \Lambda _{N{a^ + }}^\infty + \Lambda _{C{l^ - }}^\infty $

By putting the value of ionic conductance $\left( {\lambda _{N{a^ + }}^\infty } \right)$ and ionic conductance $\left( {\lambda _{C{l^ - }}^\infty } \right)$ at infinite dilution, we will get

Equivalent conductance at infinite dilution$\left( {\Lambda _{eq}^c} \right)$

= 43.0 + 65

= 108

To find the degree of dissociation $\left( \alpha \right)$

We know, degree of dissociation is the ratio of equivalent conductance at concentration to the equivalent conductances at infinite dilution i.e.

Degree of dissociation\[\left( \alpha \right) = \dfrac{{\Lambda _{eq}^c}}{{\Lambda _{eq}^\infty }}\]

By putting the value, we will get

Degree of dissociation $\left( \alpha \right) = \dfrac{{92}}{{108}}$ = 0.85