How does $H^{1,1}(X,\mathbb Z)$ look like in $H^{1,1}(X,\mathbb R)$?

Question

Let $X$ be a compact Kahler manifold. Let $c_1:Pic(X)\rightarrow H^2(X,\mathbb Z)$ and $NS(X) = Im(c_1)$ and $H^{1,1}(X,\mathbb Z) = Im(H^2(X,\mathbb Z)\rightarrow H^2(X,\mathbb C))\cap H^{1,1}(X)$.

My question is:

1. Intuitively, how does $H^{1,1}(X,\mathbb Z)$ look like in $H^{1,1}(X,\mathbb R)$? Is it a complete lattice?

2. In Demailly's 'Analytic Methods of Algebraic Geometry', he defines the algebraic class of Kahler form as $\mathcal K_{NS} = \mathcal K\cap NS_\mathbb R(X)$ when $X$ is projective. But Lefschetz (1,1)-theorem tells us $Pic(X)\rightarrow H^{1,1}(X,\mathbb Z)$ is surjective. So, why don't we have $\mathcal K\subseteq NS_{\mathbb R}(X)$?