The conductivity of a saturated solution of ${BaSO}_4$ is $3.06\times {10}^{-6}{ohm}^{-1}{cm}^{-1}$ and its molar conductance is $1.53{ohm}^{-1}{cm}^{-1}{mol}^{-1}$ the $K_{sp}$ of ${BaSO}_4$ will be ?

The correct option is B $4\times {10}^{-6}$

${\Lambda }_m=K\times \frac{1000}{M}$

$M=K\times \frac{1000}{{\Lambda }_m}$

$M=\frac{3.06\times {10}^{-6}{ohm}^{-1}{cm}^{-1}\times 1000}{1.53{ohm}^{-1}{cm}^{-1}{mol}^{-1}}$

$M=2\times {10}^{-3}$

$M=S=2\times {10}^{-3}$

$K_{sp}=\left[{Ba}^{2+}\right]\left[{SO}_4^{2-}\right]$

$K_{sp}=S\times S=S^2$

$K_{sp}=4\times {10}^{-6}$