9. The number of spatial quadrilaterals with vertices at the vertices and midpoints of the edges of a tetrahedron is ( ).
(A) 141
(B) 144
(C) 423
(D) 432
9. C.

A tetrahedron has four vertices, six edges, and each edge has a midpoint, making a total of ten points. The number of ways to choose four non-coplanar points from these ten points is as follows:
(1) The number of ways to choose four points that are all vertices is only 1;
(2) The number of ways to choose four points with three vertices is $4 \times 3=12$;
(3) The number of ways to choose four points with two vertices is $6\left(\mathrm{C}_{5}^{2}-2\right)=48$;
(4) The number of ways to choose four points with one vertex is $4\left(C_{6}^{3}-3\right)=68$;
(5) The number of ways to choose four points with no vertices is $\mathrm{C}_{6}^{4}-3=12$.

In summary, the number of ways to choose four non-coplanar points from these ten points is $1+12+48+68+12=141$.

Since each set of four non-coplanar points can form three different spatial quadrilaterals, the total number of spatial quadrilaterals with these ten points as vertices is $141 \times 3=423$.