The freezing point of a solution that contains 10 g urea in 100 g water is `(K_f for H_2O = 1.86°C m^(-1))`

Options

A. -5.1°C
B. -2.5°C
C. -3.6°C
D. -3.1°C

Answer

To solve the problem of finding the freezing point of a solution containing 10 g of urea in 100 g of water, we will use the formula for freezing point depression:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \(\Delta T_f\) = freezing point depression
- \(K_f\) = freezing point depression constant for water (1.86 °C kg/mol)
- \(m\) = molality of the solution

### Step 1: Calculate the number of moles of urea
First, we need to find the number of moles of urea. The molar mass of urea (NH₂CONH₂) is approximately 60 g/mol.

\[ \text{Number of moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{10 \, \text{g}}{60 \, \text{g/mol}} = \frac{1}{6} \, \text{mol} \approx 0.1667 \, \text{mol} \]

### Step 2: Calculate the mass of the solvent in kg
The mass of water (solvent) is given as 100 g. We need to convert this to kg for the molality calculation.

\[ \text{Mass of water in kg} = \frac{100 \, \text{g}}{1000} = 0.1 \, \text{kg} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

\[ m = \frac{\text{moles of urea}}{\text{mass of water in kg}} = \frac{0.1667 \, \text{mol}}{0.1 \, \text{kg}} = 1.667 \, \text{mol/kg} \]

### Step 4: Calculate the freezing point depression
Now we can calculate the freezing point depression using the \(K_f\) value.

\[ \Delta T_f = K_f \cdot m = 1.86 \, °C \cdot 1.667 \, \text{mol/kg} \approx 3.1 \, °C \]

### Step 5: Determine the new freezing point of the solution
The normal freezing point of water is 0 °C. The freezing point of the solution will be lower than this by the amount of \(\Delta T_f\).

\[ T_f = 0 \, °C - \Delta T_f = 0 \, °C - 3.1 \, °C = -3.1 \, °C \]

### Final Answer
The freezing point of the solution is \(-3.1 \, °C\).

---