The density of $\ce{CuCl}$ is given – $x\ \mathrm{g/cm^3}$
The crystal structure is assumed to be fcc.
My teacher is claiming the we can apply the formula
$$\rho=\frac{Z \cdot M}{a^3 \cdot N_\mathrm A}$$ And he took $M = 35.5\ \mathrm{g\ mol^{-1}}+63.5\ \mathrm{g\ mol^{-1}} = 99\ \mathrm{g\ mol^{-1}}$, $Z = 4$ as it is fcc. But my question is how can we take $M = 99\ \mathrm{g\ mol^{-1}}$? It means every particle in the fcc has mass $99\ \mathrm u$ doesn't it?
Well, yes, depending on the detail you want to go into. You can find a representation of the crystal structure of CuCl at, for example, http://www.webelements.com/compounds/copper/copper_chloride.html - you will see that the Cu atoms are indeed a simple fcc lattice. The Cl atoms are on sites that are an interpenetrating fcc lattice. The end result is the zinc blende (ZnS) lattice prototype, very familiar to semiconductor folks as the crystal structure for GaAs an associated III-V compounds. If all sites were occupied by the same atom, this is known as diamond cubic, commonly seen in diamond, Si, and Ge (amongst others).
But, yes, one way to consider the crystal (particularly for your purposes here) is an fcc crystal with a Cu-Cl basis. Just don't expect the x-ray diffraction to correspond to fcc.
