An aqueous solution contain 3% and 1.8% by mass urea and glucose respectively. What is the freezing point of solution ? (`K_(f)=1.86^(@)C//m`)

Options

A. `-1.172^(@)C`
B. `-2.27^(@)C`
C. `-1.5^(@)C`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We have an aqueous solution containing:
- 3% by mass of urea
- 1.8% by mass of glucose
- The freezing point depression constant \( K_f = 1.86 \, ^\circ C/m \)

### Step 2: Calculate the mass of solutes
For a 100 g solution:
- Mass of urea = 3 g
- Mass of glucose = 1.8 g

Total mass of solutes = 3 g (urea) + 1.8 g (glucose) = 4.8 g

### Step 3: Calculate the mass of the solvent (water)
Mass of solvent = Total mass of solution - Total mass of solutes
= 100 g - 4.8 g = 95.2 g

### Step 4: Calculate the molality of each solute
**Molar mass of urea (NH₂CONH₂)** = 60 g/mol
**Molar mass of glucose (C₆H₁₂O₆)** = 180 g/mol

**Molality of urea**:
\[ \text{Molality (m)} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{mass of solvent (kg)}} \]
\[ \text{Molality of urea} = \frac{3 \, \text{g}}{60 \, \text{g/mol} \times 0.0952 \, \text{kg}} = \frac{3}{60 \times 0.0952} \approx 0.525 \, m \]

**Molality of glucose**:
\[ \text{Molality of glucose} = \frac{1.8 \, \text{g}}{180 \, \text{g/mol} \times 0.0952 \, \text{kg}} = \frac{1.8}{180 \times 0.0952} \approx 0.105 \, m \]

### Step 5: Calculate the total molality of the solution
Total molality \( m_{total} = m_{urea} + m_{glucose} \)
\[ m_{total} = 0.525 + 0.105 = 0.630 \, m \]

### Step 6: Calculate the freezing point depression (\( \Delta T_f \))
Using the formula:
\[ \Delta T_f = K_f \times m_{total} \]
\[ \Delta T_f = 1.86 \, ^\circ C/m \times 0.630 \, m \approx 1.1738 \, ^\circ C \]

### Step 7: Calculate the freezing point of the solution
The freezing point of the solution is given by:
\[ T_f = T_f^0 - \Delta T_f \]
Since the freezing point of pure water \( T_f^0 \) is 0 °C:
\[ T_f = 0 - 1.1738 \approx -1.1738 \, ^\circ C \]

### Final Answer:
The freezing point of the solution is approximately \(-1.17 \, ^\circ C\).

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