An atomic substance A of molar mass 12 g mol$$^{-1}$$ has a cubic crystal structure with edge length of 300 pm. The no. of atoms present in one unit cell of A is _______ (Nearest integer)
Given the density of A is 3.0 g m m$$^{-1}$$ and NA$$_A$$ = $$6.02 \times 10^{23}$$ mol$$^{-1}$$

For a cubic crystal structure, the number of atoms per unit cell is given by:

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

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

$$\rho = 3.0$$ g cm$$^{-3}$$

$$a = 300$$ pm = $$3 \times 10^{-8}$$ cm

$$M = 12$$ g mol$$^{-1}$$

$$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$

Substituting the known values:

$$a^3 = (3 \times 10^{-8})^3 = 27 \times 10^{-24} \text{ cm}^3$$

$$Z = \frac{3.0 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{12}$$

$$Z = \frac{3.0 \times 27 \times 0.602}{12}$$

$$Z = \frac{48.762}{12} = 4.06 \approx 4$$

The number of atoms present in one unit cell is $$4$$ (corresponding to an FCC unit cell).