Suppose $A$ is a unital $C^*$ algebra,$\pi:A\to B(H)$ is a representation of $A$.Then the von Neumann algebra generated by $\pi(A)$ is equal to $\pi(A)^{"}$.
Is the weak$*$ closure of $\pi(A)$ equal to $\pi(A)^{"}$?When $A$ is non-unital,Is the von Neumann algebra generated by $\pi(A)$ also equal to $\pi(A)^{"}$?
Yes, unless your representation is degenerate, meaning that $\pi(A)H$ is not dense. For an easy example of what the problem is consider $A=c_0$ and $\pi:A\to B(\ell^2(\mathbb N)\oplus \ell^2(\mathbb N))$ be given by $\pi(a)(x,y)=(ax,0)$. This is degenerate because $\pi(A)H\subset \ell^2(\mathbb N)\oplus 0$. Now you have $$ \pi(A)''=\ell^\infty(\mathbb N)\oplus B(\ell^2(\mathbb N)), $$ because anything of the form $(0,y)$ commutes with everything coming from $\pi$. The von Neumann algebra generated by $\pi(A)=c_0 \oplus 0$ is arguably $\ell^\infty(\mathbb N)\oplus 0$, which does not have the same identity as $B(H)$.