To solve the problem of finding the freezing point of a solution containing 3% urea and 1.8% glucose by mass, we will follow these steps:
### Step 1: Understand the Problem
We need to calculate the freezing point depression (ΔTf) of the solution using the formula:
\[ \Delta T_f = K_f \cdot m \]
where \( K_f \) is the freezing point depression constant and \( m \) is the molality of the solution.
### Step 2: Calculate the Mass of Solvent
Assuming we have 100 grams of the solution:
- Mass of urea = 3 grams
- Mass of glucose = 1.8 grams
- Mass of solvent (water) = Total mass - Mass of solute
\[ \text{Mass of solvent} = 100 \, \text{g} - (3 \, \text{g} + 1.8 \, \text{g}) = 95.2 \, \text{g} \]
### Step 3: Calculate the Moles of Urea and Glucose
1. **Molecular mass of urea (NH₂CONH₂)**:
- N: 14 g/mol (2 nitrogen atoms)
- H: 1 g/mol (4 hydrogen atoms)
- C: 12 g/mol (1 carbon atom)
- O: 16 g/mol (1 oxygen atom)
- Total = \( 2(14) + 4(1) + 12 + 16 = 60 \, \text{g/mol} \)
Moles of urea:
\[ \text{Moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \, \text{g}}{60 \, \text{g/mol}} = 0.05 \, \text{mol} \]
2. **Molecular mass of glucose (C₆H₁₂O₆)**:
- C: 12 g/mol (6 carbon atoms)
- H: 1 g/mol (12 hydrogen atoms)
- O: 16 g/mol (6 oxygen atoms)
- Total = \( 6(12) + 12(1) + 6(16) = 180 \, \text{g/mol} \)
Moles of glucose:
\[ \text{Moles of glucose} = \frac{1.8 \, \text{g}}{180 \, \text{g/mol}} = 0.01 \, \text{mol} \]
### Step 4: Calculate the Total Moles of Solute
Total moles of solute:
\[ \text{Total moles} = \text{Moles of urea} + \text{Moles of glucose} = 0.05 + 0.01 = 0.06 \, \text{mol} \]
### Step 5: Calculate the Molality of the Solution
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
\[ m = \frac{0.06 \, \text{mol}}{0.0952 \, \text{kg}} \approx 0.630 \, \text{mol/kg} \]
### Step 6: Calculate the Freezing Point Depression
Using the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \]
Given \( K_f = 1.86 \, ^\circ C/m \):
\[ \Delta T_f = 1.86 \, ^\circ C/m \cdot 0.630 \, m \approx 1.1698 \, ^\circ C \]
### Step 7: Calculate the New Freezing Point
The freezing point of pure water is \( 0 \, ^\circ C \). Therefore, the freezing point of the solution is:
\[ T_f = 0 - \Delta T_f = 0 - 1.1698 \approx -1.17 \, ^\circ C \]
### Final Answer
The freezing point of the solution is approximately:
\[ T_f \approx -1.17 \, ^\circ C \]
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