A very small amount of a non-volatile solute (that does not dissociate) is dissolved in `56.8 cm^(3)` of benzene (density `0.889 g cm^(3))`. At room temperature, vapour pressure of this solution is `98.88 mm Hg` while that of benzene is `100 mm Hg` . Find the molality of this solution. If the freezing temperature of this solution is `0.73` degree lower than that of benzene, what is the value of molal the freezing point depression constant of benzene?

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript while ensuring clarity in each step.

### Step 1: Calculate the mole fraction of benzene.
Given:
- Vapor pressure of the solution (P1) = 98.88 mm Hg
- Vapor pressure of pure benzene (P0) = 100 mm Hg

Using Raoult's Law:
\[ X_1 = \frac{P_1}{P_0} = \frac{98.88}{100} = 0.9888 \]

### Step 2: Calculate the moles of benzene.
Assuming we have 1 mole of the total solution (solvent + solute):
- Mole fraction of benzene (X1) = moles of benzene (n1) / total moles (n_total)

Since n_total = 1 mole:
\[ n_1 = X_1 = 0.9888 \text{ moles of benzene} \]

### Step 3: Calculate the moles of solute.
Using the relationship:
\[ n_2 = n_{total} - n_1 = 1 - 0.9888 = 0.0112 \text{ moles of solute} \]

### Step 4: Calculate the mass of benzene.
Given the density of benzene = 0.889 g/cm³ and volume = 56.8 cm³:
\[ \text{Mass of benzene (W1)} = \text{Density} \times \text{Volume} = 0.889 \, \text{g/cm}^3 \times 56.8 \, \text{cm}^3 = 50.5 \, \text{g} \]

Convert grams to kilograms:
\[ W1 = 50.5 \, \text{g} = 0.0505 \, \text{kg} \]

### Step 5: Calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n_2}{W1} = \frac{0.0112 \, \text{moles}}{0.0505 \, \text{kg}} = 0.2218 \, \text{mol/kg} \]

### Step 6: Calculate the freezing point depression constant (Kf).
Given that the freezing point depression (ΔTf) is 0.73 °C:
Using the formula:
\[ \Delta T_f = K_f \times m \]
Rearranging to find Kf:
\[ K_f = \frac{\Delta T_f}{m} = \frac{0.73}{0.2218} = 3.29 \, \text{°C kg/mol} \]

### Final Answers:
- **Molality of the solution**: 0.2218 mol/kg
- **Freezing point depression constant (Kf)**: 3.29 °C kg/mol