Specific conductance of 0.1 M Nitric acid is $6.3\times {10}^{-2}{\mathrm{ohm}}^{-1}{\mathrm{cm}}^{-1}$ The molar conductance of the solution is 

$$\begin{gathered} specific conductivity\left(k\right)\hspace{0.17em}= 6.3 \times 10-2 {\Omega }^{-1}.c{m}^{-1}. \\ Concentration \left(C\right)\hspace{0.17em}=0.1M \\ Molar conductivity \left({\wedge }_m\right)=\frac{1000\times k}{C} \\ =\frac{1000\times 6.3\times {10}^{-2}}{0.1} \\ =630{\Omega }^{-1}.c{m}^2.mo{l}^{-1}. \end{gathered}$$