Equivalent conductivity of $$\mathrm{BaCl_2}$$, $$\mathrm{H_2SO_4}$$ and $$\mathrm{HCl}$$ are $$x_1,\ x_2$$ and $$x_3\ \mathrm{S\,cm^{-1}\,eq^{-1}}$$ at infinite dilution. If the conductivity of saturated $$\mathrm{BaSO_4}$$ solution is $$x\ \mathrm{S\,cm^{-1}}$$, then $$K_{sp}$$ of $$\mathrm{BaSO_4}$$ is:

$$\Lambda_{\mathrm{eq}}^\infty(\mathrm{BaSO_4})=2\Lambda_{\mathrm{eq}}^\infty(\mathrm{BaSO_4})$$

$$\Lambda_{\mathrm{eq}}^\infty(\mathrm{BaSO_4})=\Lambda_{\mathrm{eq}}^\infty(\mathrm{Ba^{2+}})+\Lambda_{\mathrm{eq}}^\infty(\mathrm{SO_4^{2-}})$$

$$\Lambda_{\mathrm{eq}}^\infty(\mathrm{BaCl_2})+\Lambda_{\mathrm{eq}}^\infty(\mathrm{H_2SO_4})-\Lambda_{\mathrm{eq}}^\infty(\mathrm{HCl})$$

$$\Lambda_{\mathrm{eq}}^\infty(\mathrm{BaSO_4})=x_1+x_2-x_3$$
$$\Lambda_m^\infty=2(x_1+x_2-x_3)$$

For a sparingly soluble salt:
$$\Lambda_m=\frac{\kappa}{M}\times1000$$
$$M=\frac{x}{2(x_1+x_2-x_3)}\times1000$$
$$M=\frac{500x}{x_1+x_2-x_3}$$
$$K_{sp}=M^2$$

$$\boxed{K_{sp}=\frac{2.5\times10^5x^2}{(x_1+x_2-x_3)^2}}$$

Correct option: (C)