hrm{cm}^{-3} \, \mathrm{mol}^{-1} $$, follow these steps:

Understand the Relation: Density ($\varrho$) is given by the formula:

$$ \varrho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} $$

For an element forming an FCC structure, the number of atoms per unit cell is 4. Using the molar mass ($M$) and Avogadro's number ($N_{\mathrm{A}}$), the mass of atoms in a unit cell is:

$$ \text{mass} = \frac{4 \times M}{N_{\mathrm{A}}} $$

Equating it to the provided density expression:

$$ \frac{4 \times M}{N_{\mathrm{A}}} = \varrho \times V_{\text{cell}} $$

Insert the Given Values:

Molar mass, $M = 63.5 \, \mathrm{g/mol}$

Avogadro's number, $N_{\mathrm{A}} = 6.022 \times 10^{23} \, \mathrm{mol}^{-1}$

Substitute $\varrho \times \mathrm{N}_{\mathrm{A}} = 5.5 \times 10^{24} \, \mathrm{g} \, \mathrm{cm}^{-3} \, \mathrm{mol}^{-1}$

The actual density $\varrho$ is:

$$ \varrho = \frac{5.5 \times 10^{24}}{6.022 \times 10^{23}} $$

Calculate the Volume ($V_{\text{cell}}$):

Substitute the expression for $\varrho$ back:

$$ \frac{4 \times 63.5}{6.022 \times 10^{23}} = \left( \frac{5.5 \times 10^{24}}{6.022 \times 10^{23}} \right) \times V_{\text{cell}} $$

Solving for $V_{\text{cell}}$ gives:

$$ V_{\text{cell}} = \frac{4 \times 63.5}{5.5 \times 10^{24}} \times 6.022 \times 10^{23} $$

$$ V_{\text{cell}} = \frac{254}{5.5} \times 0.1003 \times 10^{-1} $$

$$ V_{\text{cell}} = 4.618 \times 10^{-23} \, \mathrm{cm}^3 $$

Thus, the volume of the unit cell is $\boxed{4.618 \times 10^{-23} \mathrm{~cm}^3}$, which corresponds to Option D.

Solid State

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