Question

An element with molar mass 27 g/mol forms a cubic unit cell with edge length 405 pm. If density of the crystal is g cm\(^{-3}\), identify the type of unit cell. Derive the equation of Raoult’s law for binary solution containing non-volatile solute.

Hint:

For unit cell calculations, ensure all units are consistent before plugging them into the formula (e.g., convert pm to cm). For Raoult's law, remember that a non-volatile solute lowers the vapour pressure of the solvent because it reduces the fraction of solvent molecules at the surface available to escape into the vapour phase.

Answer 1

Identifying the Type of Unit Cell
Step 1: List the given information.
- Molar mass (M) = 27 g/mol
- Edge length = 405 pm = 405 \( \times \) 10\(^{-10}\) cm = 4.05 \( \times \) 10\(^{-8}\) cm
- Density (\(\rho\)) = g cm\(^{-3}\)
- Avogadro's number (N\(_A\)) = 6.022 \( \times \) 10\(^{23}\) mol\(^{-1}\)

Step 2: Use the formula for the density of a unit cell.
The density is related to the molar mass and unit cell dimensions by the formula:
\[ \rho = \frac{z \cdot M}{a^3 \cdot N_A} \]
where 'z' is the number of atoms per unit cell.
Step 3: Rearrange the formula to solve for z.
\[ z = \frac{\rho \cdot a^3 \cdot N_A}{M} \]
Step 4: Substitute the values and calculate z.
\[ z = \frac{( \, \text{g cm}^{-3}) \cdot (4.05 \times 10^{-8} \, \text{cm})^3 \cdot (6.022 \times 10^{23} \, \text{mol}^{-1})}{27 \, \text{g mol}^{-1}} \]
\[ z = \frac{() \cdot (66.43 \times 10^{-24}) \cdot (6.022 \times 10^{23})}{27} \] \[ z = \frac{108.0}{27} = 4 \]
Step 5: Identify the unit cell type.
Since the number of atoms per unit cell (z) is 4, the unit cell is a face-centred cubic (FCC) unit cell.
Derivation of Raoult's Law for a Solution with a Non-Volatile Solute
Step 1: State Raoult's Law.
Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. For a component 1, \( P_1 \propto x_1 \), or \( P_1 = P_1^0 x_1 \), where \( P_1^0 \) is the vapour pressure of the pure component.
Step 2: Consider a binary solution with a non-volatile solute.
Let the solvent be component 1 and the non-volatile solute be component 2. The vapour pressure of the solution will only be due to the solvent, as the solute does not contribute to the vapour.
Therefore, the vapour pressure of the solution (P\(_\text{solution}\)) is equal to the partial vapour pressure of the solvent (P\(_1\)). \[ P_\text{solution} = P_1 = P_1^0 x_1 \quad \cdots (1) \]
Step 3: Define the lowering of vapour pressure.
The lowering of vapour pressure (\(\Delta P\)) is the difference between the vapour pressure of the pure solvent (\(P_1^0\)) and the vapour pressure of the solution (\(P_\text{solution}\)).
\[ \Delta P = P_1^0 - P_\text{solution} \]
Substituting equation (1):
\[ \Delta P = P_1^0 - P_1^0 x_1 = P_1^0 (1 - x_1) \]
Step 4: Relate to the mole fraction of the solute.
For a binary solution, the sum of the mole fractions is 1: \( x_1 + x_2 = 1 \). Therefore, \( 1 - x_1 = x_2 \).
Substituting this into the equation for \( \Delta P \): \[ \Delta P = P_1^0 x_2 \]
Step 5: Derive the expression for relative lowering of vapour pressure.
The relative lowering of vapour pressure is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent.
\[ \frac{\Delta P}{P_1^0} = \frac{P_1^0 x_2}{P_1^0} \] \[ \frac{\Delta P}{P_1^0} = x_2 \] This equation, \( \frac{P_1^0 - P_\text{solution}}{P_1^0} = x_2 \), is the mathematical expression of Raoult's law for a solution containing a non-volatile solute. It shows that the relative lowering of vapour pressure is equal to the mole fraction of the solute.