Given that ${\wedge }_{\mathrm{m}}^{\infty }=133.4\left({\mathrm{AgNO}}_3\right)$${\wedge }_{\mathrm{m}}^{\infty }=149.9\left(\mathrm{KCl}\right);{\wedge }_{\mathrm{m}}^{\infty }=144.9{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}\left({\mathrm{KNO}}_3\right)$  the molar conductivity at infinite dilution for AgCl is:

$\begin{array}{l}{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{AgNO}}_3\right)={\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{Ag}}^{+}\right)+{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{NO}}_3^{-}\right)=133.4{\mathrm{sm}}^2{\mathrm{mole}}^{-1}..........\left(1\right)\\ {\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left(\mathrm{KCl}\right)={\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{K}}^{+}\right)+{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{Cl}}_{ }^{-}\right)=149.9{\mathrm{sm}}^2{\mathrm{mole}}^{-1}............\left(2\right)\\ {\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{KNO}}_3\right)={\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{K}}^{+}\right)+{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left({\mathrm{NO}}_3^{-}\right)=144.9{\mathrm{sm}}^2{\mathrm{mole}}^{-1} ............\left(3\right)\\ {\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{\infty }}\left(\mathrm{AgCl}\right)=\mathrm{Eq}\left(1\right)+\mathrm{Eq}\left(2\right)-\mathrm{Eq}\left(3\right)\end{array}$

= (133.4 + 149.9 – 144.9)${\mathrm{Scm}}^2 {\mathrm{mole}}^{–1}$
        = 138.4 ${\mathrm{Scm}}^2 {\mathrm{mole}}^{–1}$