Question

The specific conductivity of a saturated $AgCl$ solution is $1.26\times 10^{-6}Sc{m}^{-1}$ If $K_{sp}$ of $AgCl$ is $x\times 10^{-12},$ then value of ${}^{\prime }{x}^{\prime }$ is ?

Answer 1

Answer: 81

${\Lambda }_m^{\infty }(AgCl)=62+78=140$
$140=\frac{1.26\times 10^{-6}\times 1000}{s}$
$s=9\times 10^{-6}$
$K_{sp}=s^2=81\times 10^{-12}$