The question is in subject.
Update: See Andreas Thom's answer.
The question is in subject. Update: See Andreas Thom's answer. 


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. 


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 $\mathbb C$. Note that I have never assumed that the metric, the vector space structure, and the additional operations interact in any way whatsoever. 

