The molar conductivity of  $AgN{O}_3,NaCN,andNaN{O}_3$ at infinite dilution are 116.5, 110.3, and 105.2 mho  $c{m}^{2}mo{l}^{-1}$ respectively. In the same unit, the molar conductivity of  $AgCN$ is 

Given that, 
The molar conductivity of  solutions at infinite dilution

$\begin{array}{l}{\wedge }_{AgN{O}_3}^{\infty }=116.5mhoc{m}^{2}mo{l}^{-1}\\ {\wedge }_{NaCN}^{\infty }=110.3mhoc{m}^{2}mo{l}^{-1}\\ {\wedge }_{NaN{O}_3}^{\infty }=105.2mhoc{m}^{2}mo{l}^{-1}\\ {\wedge }_{AgCN}^{\infty }={\wedge }_{AgN{O}_3}^{\infty }+{\wedge }_{NaCN}^{\infty }-{\wedge }_{NaN{O}_3}^{\infty }\\ =116.5+110.3-105.2\\ =121.6mho\text{\hspace{0.17em}\hspace{0.17em}}c{m}^{2}mo{l}^{-1}\end{array}$