A very small amount of non-volatile solute, non-electrolyte is dissolved in `50 cm^(3)` of benzene `(d=0.9 g cm^(-3))`. At room temperature, vapour prssure of this solution is 99 mm Hg while that of benzene is 100 Hg. If `DeltaT_(f)` is the `0.73` What is value of `K_(f)?`

Answer

To solve the problem, we need to find the value of the cryoscopic constant \( K_f \) for benzene using the given data. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the weight of the solvent (benzene)
Given:
- Volume of benzene = \( 50 \, \text{cm}^3 \)
- Density of benzene = \( 0.9 \, \text{g/cm}^3 \)

Using the formula for weight:
\[ \text{Weight} = \text{Density} \times \text{Volume} \]
\[ \text{Weight of benzene} = 0.9 \, \text{g/cm}^3 \times 50 \, \text{cm}^3 = 45 \, \text{g} \]

### Step 2: Convert the weight of the solvent to kilograms
Since molality is expressed in kilograms, we convert grams to kilograms:
\[ \text{Weight of benzene in kg} = \frac{45 \, \text{g}}{1000} = 0.045 \, \text{kg} \]

### Step 3: Calculate the relative lowering of vapor pressure
Given:
- Vapor pressure of pure benzene \( P_0 = 100 \, \text{mm Hg} \)
- Vapor pressure of the solution \( P_s = 99 \, \text{mm Hg} \)

The relative lowering of vapor pressure is given by:
\[ \text{Relative lowering} = \frac{P_0 - P_s}{P_0} = \frac{100 - 99}{100} = \frac{1}{100} = 0.01 \]

### Step 4: Relate the relative lowering of vapor pressure to mole fraction of solute
The relative lowering of vapor pressure is equal to the mole fraction of the solute (\( X_B \)):
\[ X_B = \frac{n_B}{n_A + n_B} \approx \frac{n_B}{n_A} \quad (\text{since } n_B \text{ is very small}) \]
Where:
- \( n_A \) = moles of benzene (solvent)
- \( n_B \) = moles of solute

### Step 5: Calculate the number of moles of benzene
The molecular mass of benzene \( C_6H_6 \) is:
\[ \text{Molar mass of benzene} = 6 \times 12 + 6 \times 1 = 78 \, \text{g/mol} \]
Now, calculate the moles of benzene:
\[ n_A = \frac{\text{Weight of benzene}}{\text{Molar mass of benzene}} = \frac{45 \, \text{g}}{78 \, \text{g/mol}} \approx 0.5769 \, \text{mol} \]

### Step 6: Calculate the moles of solute using the mole fraction
From the relative lowering of vapor pressure:
\[ 0.01 = \frac{n_B}{n_A} \implies n_B = 0.01 \times n_A = 0.01 \times 0.5769 \approx 0.005769 \, \text{mol} \]

### Step 7: Calculate the molality of the solution
Molality (\( m \)) is defined as:
\[ m = \frac{n_B}{\text{Weight of solvent in kg}} = \frac{0.005769 \, \text{mol}}{0.045 \, \text{kg}} \approx 0.1283 \, \text{mol/kg} \]

### Step 8: Use the freezing point depression formula to find \( K_f \)
Given:
- \( \Delta T_f = 0.73 \, \text{°C} \)

The freezing point depression formula is:
\[ \Delta T_f = K_f \times m \]
Rearranging for \( K_f \):
\[ K_f = \frac{\Delta T_f}{m} = \frac{0.73 \, \text{°C}}{0.1283 \, \text{mol/kg}} \approx 5.69 \, \text{°C kg/mol} \]

### Final Answer:
\[ K_f \approx 5.69 \, \text{°C kg/mol} \]