1.00 g of a non-electrolyte solute (molar mass 250g `mol^(–1)`) was dissolved in 51.2 g of benzene. If the freezing point depression constant `K_(f)` of benzene is 5.12 K kg `mol^(–1)`, the freezing point of benzene will be lowered by:-

Options

A. 0.4 K
B. 0.3 K
C. 0.5 K
D. 0.2 K

Explanation

To solve the problem of calculating the freezing point depression of benzene when a non-electrolyte solute is dissolved in it, we will follow these steps:

### Step 1: Identify the given data
- Mass of solute (W) = 1.00 g
- Molar mass of solute (M) = 250 g/mol
- Mass of solvent (benzene) = 51.2 g
- Freezing point depression constant (K_f) of benzene = 5.12 K kg/mol

### Step 2: Convert the mass of the solvent from grams to kilograms
To calculate molality, we need the mass of the solvent in kilograms:
- Mass of benzene in kg = 51.2 g × (1 kg / 1000 g) = 0.0512 kg

### Step 3: Calculate the number of moles of solute
Using the formula for moles:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{1.00 \text{ g}}{250 \text{ g/mol}} = 0.004 \text{ mol} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.004 \text{ mol}}{0.0512 \text{ kg}} \approx 0.078125 \text{ mol/kg} \]

### Step 5: Calculate the freezing point depression (ΔT_f)
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \times \text{molality} \]
Substituting the values:
\[ \Delta T_f = 5.12 \text{ K kg/mol} \times 0.078125 \text{ mol/kg} \approx 0.4 \text{ K} \]

### Conclusion
The freezing point of benzene will be lowered by approximately **0.4 K**.

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