Find the `K_(f) ` if 6 g of urea is dissolved in 0.1 `dm^(3)` of water and it corresponds to `0.15 ""^(@) C ` in `Delta T` ( Molacular weight of urea = 60 g `mol ^(-1) `)

Options

A. `0.015`
B. `0.15`
C. `0.30`
D. `0.030`

Answer

To find the freezing point depression constant \( K_f \) for urea dissolved in water, we can follow these steps:

### Step 1: Understand the given data
- Mass of urea (solute) = 6 g
- Volume of water (solvent) = 0.1 dm³ (which is equivalent to 0.1 L)
- Change in freezing point \( \Delta T_f \) = 0.15 °C
- Molar mass of urea = 60 g/mol

### Step 2: Calculate the number of moles of urea
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles of urea} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Calculate the mass of the solvent in kg
Since we have 0.1 dm³ of water and the density of water is approximately 1 kg/L:
\[ \text{Mass of water} = 0.1 \, \text{L} \times 1 \, \text{kg/L} = 0.1 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{0.1 \, \text{mol}}{0.1 \, \text{kg}} = 1 \, \text{mol/kg} \]

### Step 5: Use the freezing point depression formula
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = change in freezing point = 0.15 °C
- \( i \) = van 't Hoff factor (for urea, which does not dissociate, \( i = 1 \))
- \( K_f \) = freezing point depression constant (which we need to find)
- \( m \) = molality = 1 mol/kg

### Step 6: Substitute the known values into the formula
\[ 0.15 = 1 \cdot K_f \cdot 1 \]
This simplifies to:
\[ K_f = 0.15 \, \text{°C kg/mol} \]

### Conclusion
The value of \( K_f \) is **0.15 °C kg/mol**.

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