Metal M crystallizes into a FCC lattice with the edge length of $$4.0 \times 10^{-8}$$ cm. The atomic mass of the metal is _____ g/mol. (Use: $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, density of metal, $$M = 9.03$$ g cm$$^{-3}$$)

We are given that metal M crystallizes in an FCC lattice with edge length $$a = 4.0 \times 10^{-8}$$ cm, and the density $$\rho = 9.03$$ g/cm$$^3$$. We need to find the atomic mass.

For an FCC unit cell, the number of atoms per unit cell is $$Z = 4$$ (since there are 8 corner atoms each contributing $$\frac{1}{8}$$ and 6 face-centered atoms each contributing $$\frac{1}{2}$$, giving $$8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$).

The relation between density and the unit cell parameters is given by:

$$\rho = \frac{Z \times M}{a^3 \times N_A}$$

where $$M$$ is the molar mass, $$a$$ is the edge length, and $$N_A$$ is Avogadro's number.

Rearranging for $$M$$:

$$M = \frac{\rho \times a^3 \times N_A}{Z}$$

Now substituting the values:

$$a^3 = (4.0 \times 10^{-8})^3 = 64.0 \times 10^{-24} = 6.4 \times 10^{-23} \text{ cm}^3$$

So we get:

$$M = \frac{9.03 \times 6.4 \times 10^{-23} \times 6.02 \times 10^{23}}{4}$$ $$M = \frac{9.03 \times 6.4 \times 6.02}{4}$$ $$= \frac{9.03 \times 38.528}{4}$$ $$= \frac{347.9}{4}$$ $$\approx 86.98$$

Rounding to the nearest integer, we get $$M \approx 87$$ g/mol.

Hence, the correct answer is 87.