\cdot N_A = 120 \times 10^{21} \mathrm{~g/cm}^3 \mathrm{~mol}^{-1}$ and $M = 180 \mathrm{~g/mol}$.

We are supposed to find $V$, the volume of the unit cell. We can start by rearranging the formula to solve for $V$:

$$V = \frac{M}{\rho \cdot N_A}$$

Substitute the given values into the equation:

$$V = \frac{180 \mathrm{~g/mol}}{120 \times 10^{21} \mathrm{~g/cm}^3 \mathrm{~mol}^{-1}}$$

After placing the values in, you proceed with the calculation:

$$V = \frac{180}{120 \times 10^{21}} \mathrm{~cm}^3$$

$$V = \frac{3}{2} \times 10^{-21} \mathrm{~cm}^3$$

$$V = 1.5 \times 10^{-21} \mathrm{~cm}^3$$

This calculated value does not match any of the options provided, suggesting there may have been a mistake in my calculation or in interpreting the given question and options. Let's correct the oversight with the proper calculation, considering that for an FCC unit cell, there are four atoms per unit cell. The correct approach would address this relationship more precisely.

The corrected formula, taking into account the number of atoms in an FCC unit cell ($Z = 4$ for FCC), should be:

$$\rho = \frac{Z \cdot M}{N_A \cdot V}$$

So, correcting the calculation for $V$:

$$V = \frac{Z \cdot M}{\rho \cdot N_A}$$

We know $Z = 4$ for FCC, so we substitute the given values into this corrected formula:

$$V = \frac{4 \cdot 180}{120 \times 10^{21}} \mathrm{~cm}^3$$

$$V = \frac{720}{120 \times 10^{21}} \mathrm{~cm}^3$$

$$V = \frac{6}{10^{21}} \mathrm{~cm}^3$$

$$V = 6.00 \times 10^{-21} \mathrm{~cm}^3$$

Therefore, the volume of the unit cell is $6.00 \times 10^{-21} \mathrm{~cm}^3$, which corresponds to Option A.

Solid State

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