To find the molal cryoscopic constant (Kf) of benzene, we will follow these steps:
### Step 1: Calculate the change in freezing point (ΔTf)
The freezing point of pure benzene is given as 278.4 K, and the freezing point of the solution is 277.4 K.
\[
\Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 278.4 \, \text{K} - 277.4 \, \text{K} = 1.0 \, \text{K}
\]
### Step 2: Calculate the molality (m) of the solution
Given the mole fraction of acetic acid (solute) is 0.02, we can assume a total of 1 mole of solution. Therefore, the moles of solute (acetic acid) and solvent (benzene) can be calculated as follows:
- Moles of acetic acid (solute) = 0.02 moles
- Moles of benzene (solvent) = 1 - 0.02 = 0.98 moles
Next, we need to convert the moles of benzene to mass in kilograms. The molar mass of benzene (C6H6) is approximately 78 g/mol.
\[
\text{Mass of benzene} = 0.98 \, \text{moles} \times 78 \, \text{g/mol} = 76.44 \, \text{g} = 0.07644 \, \text{kg}
\]
Now, we can calculate the molality (m):
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.02 \, \text{moles}}{0.07644 \, \text{kg}} \approx 0.261 \, \text{mol/kg}
\]
### Step 3: Determine the van 't Hoff factor (i)
Acetic acid (CH3COOH) exists partly as a dimer in benzene. The degree of dimerization affects the van 't Hoff factor (i). For acetic acid, if we assume that 25% remains as monomer and 75% forms dimers, we can calculate i as follows:
\[
i = 1 \times (0.25) + 2 \times (0.75) = 0.25 + 1.5 = 1.75
\]
### Step 4: Use the freezing point depression formula
The freezing point depression formula is given by:
\[
\Delta T_f = K_f \cdot i \cdot m
\]
Substituting the known values:
\[
1.0 \, \text{K} = K_f \cdot 1.75 \cdot 0.261 \, \text{mol/kg}
\]
### Step 5: Solve for Kf
Rearranging the equation to solve for Kf:
\[
K_f = \frac{1.0 \, \text{K}}{1.75 \cdot 0.261} \approx \frac{1.0}{0.45675} \approx 2.19 \, \text{K kg/mol}
\]
### Final Answer
The molal cryoscopic constant (Kf) of benzene is approximately **2.19 K kg/mol**.
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