The \(\Lambda ^o_m\) for \(CaSO_4\) at \(298 K\), if \(\Lambda ^o_{Ca^{2+}} = 119.0\, \ Scm^2mol^{-1}\) and \(\Lambda ^o_{SO_4^{2-}} = 160.0\, \ Scm^2mol^{-1}\) will be

\(279\, \ Scm^2mol^{-1}\)

The correct answer is option 4. \(279\, \ Scm^2mol^{-1}\).

To calculate the molar conductivity at infinite dilution, \(\Lambda^o_m\), for calcium sulfate (\(CaSO_4\)) at \(298 \,K\), we use the following formula:

\(\Lambda^o_m = \Lambda^o_{Ca^{2+}} + \Lambda^o_{SO_4^{2-}}\)

Given:

\(\Lambda^o_{Ca^{2+}} = 119.0 \, Scm^2mol^{-1}\) \(\Lambda^o_{SO_4^{2-}} = 160.0 \, Scm^2mol^{-1}\)

Now, adding the contributions of both ions:

\(\Lambda^o_m = 119.0 \, Scm^2mol^{-1} + 160.0 \, Scm^2mol^{-1}\)

\(\Lambda^o_m = 279.0 \, Scm^2mol^{-1}\)

Thus, the correct answer is: 4. \(279\, Scm^2mol^{-1}\).