Is #P contained in PSPACE? It's obvious that NP $\subseteq$ #P. How about #P $\subseteq$ PSPACE? 

It strikes me as semi-obvious, since we can check whether an assignment (e.g. for SAT) is a solution in polynomial time (and hence space), and since each assignment can be checked separately, we can re-use the memory needed for this check, and thus use only a polynomial amount of space to check every assignment (plus a counter to track how many of the assignments were solutions).
For the sake of completeness, let me update this answer with a direct proof. My original answer will remain below in case anyone finds it interesting. The basic idea of the direct proof is exactly as Evgenij suggests: check whether each possible assignment is satisfiable, keep a counter of satisfiable instances, and reuse space wherever possible.

Claim: #$P \subseteq FPSPACE$

Proof: Assume #$SAT$ is #$P$-complete; then it suffices to show a polynomial-space algorithm to solve #$SAT$. Here is a linear-space, exponential-time algorithm.

Initialize two counters of $k \le n$ bits to all 0, for number of variables $k$ in the $3CNF$ formula given as input. The first counter will keep track of which assignment we are checking (1 = true, 0 = false, for each variable $x_1$ to $x_k$). The second will keep track of the number of satisfying assignments (there are at most $2^k$ of them).

We will also need enough space ($n$ bits) to write the formula, and enough space to do "scratch work" for one clause (some very small constant number of bits, since every clause will have exactly 3 literals by definition of #$SAT$).

Here is the description of the algorithm:


For all possible assignments
   For all clauses in the Boolean formula
      Check if the assignment satisfies the given clause
   If the assignment satisfied every clause
      Increment the assignment counter
Output the value in the assignment counter


Note that once we finish work on one possible assignment, we don't need any information from it again to compute whether other assignments satisfy the formula; the only information we keep is whether it was satisfied -- in the second counter. In particular, since we can just apply binary addition of 1 to the assignment we are testing in order to always get the next assignment, and since every assignment can be checked by considering only one clause at a time (and re-using the same space for computation on every clause), the dominating factor in the space requirement is simply writing the formula itself. Alternatively, we just read from an input tape and never explicitly write the formula, then in the worst case, $k = n$ and it still takes $O(n)$ space to write the possible assignments and keep track of the number of satisfying assignments. Q.E.D.



Below is my original answer:

Although this already essentially answered, let me give a proof of a different vein (straight out of Arora/Barak):

Let #$SAT_D$ = {($\phi, K$) : $\phi$ is a 3CNF formula and it has exactly $K$ satisfying assignments}

In Chapter 17, there is a proof that #$SAT$ is #$P$-complete.
In Chapter 8, there is a proof that #$SAT_D \in IP$.

Since $IP = PSPACE$, #$P \subseteq FPSPACE$.