Are AFD factors weak closures of an increasing family of full matrix algebras?

An AFD van Neumann algebra is by definition the weak closure of a *-algebra that is the union of an increasing family of finite dimensional *-algebras.

Is every AFD factor the weak closure of a *-algebra that is the union of an increasing family of full matrix algebras (i.e. all isomorphic to the algebra of complex n x n matrices for some n)?

If this is not true in general, for which AFD factors is it true? I think it is obviously true for type I and type II$_1$.

Context : I want to construct a conditional expectation on the supercommutant of AFD super ($\mathbb{Z}_2$-graded) von Neumann algebras. I can do it for super von Neumann algebras that are the weak closure of an increasing family of finite dimensional super *-algebras that are isomorphic (as ungraded algebras) to full matrix algebras. I wonder whether this covers all AFD super algebras.

Giving a non-complete answer off the top of my head. But the answer is yes.

AFD factors of types I, II$_1$, II$_\infty$, and III$_\lambda$ with $\lambda\in(0,1]$ are unique (Murray-von Neumann + Connes + Haagerup).

Type I AFD factors are either $M_n(\mathbb C)$ or $B(\ell^2(\mathbb N))$, where it is easy to see.

The hyperfinite II$_1$ factor can be realized as the closure of $\bigotimes_n M_2(\mathbb C)$ under GNS for the trace.

Writing the II$_\infty$ factor as $R\otimes B(\ell^2(\mathbb N))$ allows us to see it as the closure of an increasing union of factors of the form $M_n(\mathbb C)\otimes M_d(\mathbb C)\simeq M_{nd}(\mathbb C)$.

The AFD type III$_\lambda$ factors, with $\lambda\in(0,1)$, can be realized as the Powers factors, which are closures of $\bigotimes_n M_2(\mathbb C)$ under GNS for an appropriate state $\phi_\lambda$.

What I don't remember is what the concrete realizations of the unique III$_1$ and the many $\text{III}_0$ are.