For a saturated solution of AgCl at 25°C, specific conductance 3.41 × 10^-6 ohm^-1 cm^-1 and of water used for preparing. The solution is 1.60 × 10^-6 ohm^-1 cm^-1. If ${\mathrm{\lambda }}_{\mathrm{AgCl}}^{\mathrm{\infty }}$ = 138.3 ohm^-1 cm² eq^-1, the K_sp of AgCl is 0.17 × 10^-x M².

The value of x is

For AgCl, molarity = normality 

$\kappa =3.41\times {10}^{-6}-1.6\times {10}^{-6}=1.81\times {10}^{-6}{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{-1}$

For saturated solution of sparingly soluble salt,

${\mathrm{\lambda }}_{\mathrm{V}}={\mathrm{\lambda }}^{\mathrm{\infty }}\text{ and }\text{ solubility }=\text{ normality }$

$\begin{array}{l}\therefore {\mathrm{\lambda }}_{\mathrm{V}}={\mathrm{\lambda }}^{\mathrm{\infty }}=\frac{1000\times \mathrm{K}}{\mathrm{s}}\\ \therefore \mathrm{s}=\frac{1000\times \mathrm{K}}{{\mathrm{\lambda }}^{\mathrm{\infty }}}=\frac{1000\times 1.81\times {10}^{-6}}{138.3}=1.31\times {10}^{-5}\\ \text{ Now, }{\mathrm{K}}_{\mathrm{sp}\left(\mathrm{AgCl}\right)}={\mathrm{s}}^2={\left(1.31\times {10}^{-5}\right)}^2=0.1716\times {10}^{-9}\\ \therefore \mathrm{x}=9\end{array}$