What is Conductivity, molar ionic conductivity?

Example 46.9 (Reading the two lines)

Before equivalence each $\ce{OH-}$ added removes an $\ce{H3O+}$ ($\lambda = 349.8$) and brings in a $\ce{Na+}$ ($\lambda = 50.3$): the conductivity falls steeply. After equivalence, $\ce{Na+}$ and $\ce{OH-}$ ($\lambda = 198.3$) simply accumulate: it rises. The two lines meet at $V_E = 10.0\,\mathrm{mL}$, so the acid was $0.100 \times 10.0 / 100 = 0.0100\,\mathrm{mol}/\mathrm{L}$.