About why $\rho=b_2$ for a rational surface: this has a few ingredients. Maybe I'm giving more details than you need, but others might find them useful.
The Hodge Decomposition Theorem, which says that (for any smooth $X$) the cohomology group $H^2(X,\mathbf{C})$ decomposes as a direct sum of subspaces $H^{2,0} \oplus H^{1,1} \oplus H^{0,2}$. Each of these is a vector space of differential forms of a certain type: $H^{2,0}$ is the vector space of holomorphic 2-forms on $X$, and $H^{0,2}$ is its complex conjugate, so it has the same dimension as $H^{2,0}$.
The fact that rational varieties have no holomorphic differential forms, so in particular here $H^{2,0}=0$. Together with 1, this tells you that $H^2(X,\mathbf{C})=H^{1,1}$, the space of so-called $(1,1)$-forms.
Finally, we need the Leftschetz $(1,1)$-Theorem, which says (again in general) that an element of $H^{1,1} \cap H^2(X,\mathbf{Z})$ (an integral $(1,1)$-form) is algebraic, that is, it is the class of a divisor on $X$.
Now by 2, we know that the vector space $H^{1,1}$ is spanned by the subgroup $H^2(X,\mathbf{Z})$, so 3 tells us that $H^{1,1}$ is spanned by algebraic classes. The span of the algebraic classes is exactly the Neron-Severi group tensored with $\mathbf{C}$, so we have $NS(X) \otimes \mathbf{C} = H^{1,1}=H^2(X,\mathbf{C})$, hence $\rho=b_2$.
