help please asap

help please asap The conductivity of a saturated solution of BaSO is 3.06 x 10-6 ohm-I cm-land its molar conductance is 1.53 ohm-I cm2 mol-l. The K of BaS04 will be : (1) 4 X 10-12 (3) 2.5 (2) 2.5 x 10-9

Dear student,

According to the relation ;  ${\lambda }_m = \frac{1000\times k}{M}$
                                        1.53 = 1000 $\times$ 3.06 $\times$ 10^-6  / M 
                      Then ;  M  =  2 $\times$ 10^-3 M  ( soubility of barium sulphate ) = S mol / L
 Now  as we know that The solubility product K_sp  =  S²  ( for Barium sulphate )
                                                                                = (  2 x 10^-3 )²  =  4 x 10^-6 (Mol/l )^​2  ( option 4 )
Regards