Question

A decinormal solution of $NaCl$ has specific conductivity equal to $0.0092$. If ionic conductances of $N{a}^{+}$ and $C{l}^{-}$ ions at the same temperature are $43.0$ and $65.0oh{m}^{-1}$ respectively, calculate the degree of dissociation of $NaCl$ solution.

Answer 1

Equivalent conductance of $N/10NaCl$ solution
${\wedge }_v=Sp.conductivity\times dilution$
$=0,0092\times 10,000=92oh{m}^{-1}$
${\wedge }_{\infty }={\lambda }_{N{a}^{+}}+{\lambda }_{C{l}^{-}}$
$=43.0+65.0=108oh{m}^{-1}$
Degree of dissociation, $\alpha =\frac{{\wedge }_v}{{\wedge }_{\infty }}=\frac{92}{108}=0.85$.