Calculating conductivity of a solution given molar conductivities and concentration From my understanding the conductivity of a solution ($K$) is given by the sum of the conductivities of ions in solution:

$$K = c(K_{m,1}+K_{m,2})$$

Where $c$ is the concentration, and $k_{m1}$ and $k_{m2}$ are the respective molar conductivities of the two ions. For example, with a solution of $\pu{0.1 mol L-1}$ $\ce{NaOH}$:

\begin{align}
K &amp;= 0.1(K_{m,\ce{Na+}} + K_{m,\ce{OH-}})\\
K_{m,\ce{Na+}} &amp;= \pu{50 S cm2 mol-1}\\
K_{m,\ce{OH-}} &amp;= \pu{197.9 S cm2 mol-1}\\
K &amp;= 0.1(247.5)\\
K &amp;= \pu{24.75 S m^-1} (?)
\end{align}

However I think the units are throwing me off here as my calculations aren't making much sense - I was led to believe the conductivity was much lower. 

Could you please offer some help with these calculations?
Just be explicit with all the units, including for the molar concentration, and be sure they cancel out properly.  You have $\pu{0.1 mol L-1}$ for the solution concentration, but your $K$ values are in $\pu{S cm2 mol-1}$.

So convert your molar concentration to cubic centimeters (there are $\pu{1e3 cm3}$ in one liter) and you get $\pu{1e-4 mol cm-3}$.

Then when you multiply both values and your units cancel nicely to give you

$$\pu{1e-4 mol cm-3} \times \pu{247.5 S cm2 mol-1} = \pu{2.475e-2 S cm-1}$$