Specilic conductivicy of BaSO4 solution κBaSO4=3648×10−6ohm−1cm−1 Specific conductivity of H2O,κH2O=1.25×10−6ohm−1cm−1 Now, κBaSO4=κBaSO4 (solution) −κH2O =3648×10−6−1.25×10−6 =2.398×10−6S‌cm−1 Formula used, λm°=
1000×K
S
...(i) Here, S= solubility k=kappa (specilic conductivity of BaSO4) λm(BaSO4)=λm°(Ba2+)+λm°(SO42−) (110+136.6)ohm−1cm−1=246.6ohm−1cm−1 Put values in Eq (i) we get, 246.6ohm−1cm−1 =
1000×2.398×10−6Scm−1
S
S=9.72×10−6×233=2.266×10−3gL−1 Hence, the solubility of BaSO4 at 291K will be 2.266×10−3gL−1.