For a saturated solution of AgCl at 25°C, specific conductance is $3.55\times {10}^{-6}oh{m}^{-1}c{m}^{-1}$ and that of water used for preparing the solution is $1.60\times {10}^{-6}{\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1}$. If ${\mathrm{\lambda }}_{\mathrm{AgCl}}^{\mathrm{\infty }}=150{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$, the ${\mathrm{K}}_{\mathrm{sp}}$ of AgCl is $0.17\times {10}^{-\mathrm{x}}{\mathrm{M}}^2$. The value of x is:
 

Specific conductivity of pure AgCl = specific conductivity of saturated solution of AgCl- specific conductivity of H₂O(l)
 

$\mathrm{k}=3.55\times {10}^{-6} {\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1}-1.60\times {10}^{-6} {\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1}=1.95\times {10}^{-6}{\text{\hspace{0.17em}ohm}}^{-1}{\mathrm{cm}}^{-1}$

For saturated solution of sparingly soluble salt,

${\lambda }_V={\lambda }^{\infty }$ and solubility = normality

${\lambda }_V={\lambda }^{\infty }=\frac{1000\times k}{s}$

$\mathrm{c}=\frac{1000\times \mathrm{k}}{{{\mathrm{\lambda }}^{\mathrm{\infty }}}_{\mathrm{AgCl}}}=\frac{1000\times 1.95\times {10}^{-6}}{150}=1.31\times {10}^{-5} \mathrm{M}$

Now,
${\mathrm{K}}_{\mathrm{sp}\left(\mathrm{AgCl}\right)}={\mathrm{s}}^2={\left(1.31\times {10}^{-5}\right)}^2=0.1716\times {10}^{-9}$

X=9