If ${\mathrm{\Lambda }}_c$of ${\mathrm{NH}}_4\mathrm{OH}$ is $11.5{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$, its degree of dissociation would be

(Given ${\lambda }_{{\mathrm{NH}}_4^{+}}^{\mathrm{\infty }}=73.4{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}\text{ and }{\lambda }_{\mathrm{OH}}^{\mathrm{\infty }}=197.6{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}.$) 

degree of dissociation solution is  0.0424

$\alpha =\frac{\mathrm{\Lambda }}{{\mathrm{\Lambda }}^{\mathrm{\infty }}}=\frac{11.5}{(73.4+197.6)}=\frac{11.5}{271.0}=0.0424$.

When free ions carrying current are produced at a specific concentration and are dissociated from the solute fraction, this phenomenon is known as the degree of dissociation.