A saturated in  $AgA(K_{sp}=3\times {10}^{-14})$ and  $AgB(K_{sp}=1\times {10}^{-14})$ has conductivity of  $375\times {10}^{-10}{\text{\hspace{0.17em}\hspace{0.17em}Scm}}^{-1}$. Limiting molar conductivity of $A{g}^{+}$  and  $A^{-}$ are $60{\text{\hspace{0.17em}\hspace{0.17em}Scm}}^2{\text{\hspace{0.17em}\hspace{0.17em}mol}}^{-1}$  and  $80{\text{\hspace{0.17em}\hspace{0.17em}Scm}}^2{\text{\hspace{0.17em}\hspace{0.17em}mol}}^{-1}$ respectively. If the limiting molar conductivity of $B^{-}$  is Y $\left({\text{in\hspace{0.17em}\hspace{0.17em}Scm}}^2{\text{\hspace{0.17em}\hspace{0.17em}mol}}^{-1}\right)$ . Find out the value of $\frac{Y}{54}$ ?

$AgA\rightleftharpoons \underset{(S_1+S_2)}{A{g}^{+}}+\underset{S_1}{A^{-}}[K_{SP}=3\times {10}^{-11}]$

$AgA\rightleftharpoons \underset{(S_1+S_2)}{A{g}^{+}}+\underset{S_1}{B^{-}}[K_{SP}={10}^{-14}]$

$3\times {10}^{-14}=S_1(S_1+S_2)$

${10}^{-14}=S_2(S_1+S_2)$

$3=\frac{S_1}{S_2}$

$S_2=\frac{1}{2}\times {10}^{-7}\text{\hspace{0.17em}\hspace{0.17em}mole}/\mathcal{l}$

$S_1=\frac{3}{2}\times {10}^{-7}\text{\hspace{0.17em}\hspace{0.17em}mole}/\mathcal{l}$

$K=375\times {10}^{-10}$

$K=K_{A{g}^{+}}+K_{A^{-}}+K_{B^{-}}$

$375\times {10}^{-10}=\frac{{\lambda }_{A{g}^{+}}^{\infty }(S_1+S_2)}{1000}+\frac{{\lambda }_{A^{-}}^{\infty }\left(S_1\right)}{1000}+\frac{{\lambda }_{B^{-}}^{\infty }\left(S_2\right)}{1000}$

$375\times {10}^{-7}=60(2\times {10}^{-7})+80(\frac{3}{2}\times {10}^{-7})+{\lambda }_{B^{-}}^{\infty }\times (\frac{1}{2}\times {10}^{-7})$

${\lambda }_{B^{-}}^{\infty }=270Y=270;\frac{Y}{54}=5$