Do separable $C^*$-algebras form a set? The question is in subject. 
Update: See Andreas Thom's answer.
 A: It is not so clear what you mean.
However, every separable $C^\ast$-algebra embeds in $B(\ell^2 \mathbb N)$. Hence, the isomorphism classes of separable $C^\ast$-algebras form a set.
A: Long comment:
It should be pointed out that actually much less structure than what Andreas Thom uses in his answer is needed to show that the isomorphism classes of separable $C^*$-algebras have a set of representatives:  
The crucial fact is that
there is a set of representatives of isometry classes of separable metric spaces. 
This is essentially because separable metric spaces are of bounded size (see Komjath's comment), namely of size at most $2^{\aleph_0}$. 
Each separable metric space carries only a set of vector space structures over $\mathbb C$.
Each metric vector space over $\mathbb C$ only carries a set of binary and unary operations. 
So, we obtain a set of representatives of the isomorphism classes of separable 
$C^\ast$-algebras
without ever using the structure of $C^\ast$-algebras.  Just the fact that they are separable metric spaces with a vector space structure over $\mathbb C$ and a fixed number of binary and unary operations. 
Note that I have never assumed that the metric, the vector space structure, and the additional operations interact in any way whatsoever.
