calculate the limiting molar conductivity of CaSO4.given(molar conductivity of calcium-119.0 Scm square per mol and molar conductivity of sulphate 160.0 S cm square per mol

This can be calculated using Kohlrausch law of independent migration of ions.
According to it, the limiting molar conductivity of any strong electrolyte can be found from the summation of individual molar conductivities of the cation and the anion of that electrolyte.

Thus, 


where ${\gamma }_{+} = no. of cations provided by electrolyte on dissociation$
${\gamma }_{-} = no. of anions provided by electrolye on dssociation$
${\lambda }_{+ }^{o}and {\lambda }_{-}^o are molar conductivities of C{a}^{2+} and S{O}_4^{2-} respectively$

$$\begin{gathered} Now, CaS{O}_4 \to C{a}^{2+} + S{O}_4^{2-} \\ i.e. {\gamma }_{+} = 1 \\ {\gamma }_{-} = 1 \\ and {\lambda }_{+}^o = 119 S c{m}^2 mo{l}^{-1} \\ {\lambda }_{-}^o = 160 S c{m}^2 mo{l}^{-1} \end{gathered}$$

$$\begin{gathered} So, {\Lambda }_{m \left(CaS{O}_4\right)}^o = 1\times 119 + 1\times 160 \\ = 119 + 160 \\ = 279 S c{m}^2 mo{l}^{-1} \end{gathered}$$