Just to flesh out Bill's answer and comments thereon, we have the following facts.  Let $X,Y$ be Banach spaces and $T : X \to Y$ a bounded linear operator.

1. If $T$ has dense range then $T^*$ is injective.

Since this is a standard homework problem I'll just give a hint. Suppose $f \in Y^*$ with $T^*f=0$.  This means that $f(Tx)=0$ for every $x$, i.e. $f$ vanishes on the range of $T$...

2. Suppose further that $X$ is reflexive.  If $T$ is injective then $T^*$ has dense range.

Hint: By Hahn-Banach, to show that $T^*$ has dense range, it suffices to show that if $u \in X^{**}$ vanishes on the range of $T^*$, then $u=0$.  And by reflexivity, $u \in X^{**}$ is represented by some $x \in X$...

Note that the proofs I have in mind don't need to discuss the weak-* topology (at least not explicitly).

We cannot drop reflexivity in 2.  Consider the inclusion of $\ell^1$ into $\ell^2$.  It is injective, but its adjoint is the inclusion of $\ell^2$ into $\ell^\infty$, whose range is not dense.