The freezing point of aqueous solution that contains `5%` by mass urea. `1.0%` by mass `KCl` and `10%` by mass of glucose is:
`(K_(f) H_(2)O = 1.86 K "molality"^(-1))`

Options

A. 290.2 K
B. 285.5 K
C. 269.63 K
D. 250 K

Answer

To find the freezing point of an aqueous solution containing 5% by mass urea, 1% by mass KCl, and 10% by mass glucose, we will follow these steps:

### Step 1: Calculate the Mass of Solute and Solvent
Given that the total mass of the solution is 100 g:
- Mass of urea (W_urea) = 5 g
- Mass of KCl (W_KCl) = 1 g
- Mass of glucose (W_glucose) = 10 g

Total mass of solute = W_urea + W_KCl + W_glucose = 5 g + 1 g + 10 g = 16 g

Mass of solvent (water) = Total mass of solution - Total mass of solute = 100 g - 16 g = 84 g

### Step 2: Calculate the Molar Mass of Each Solute
- Molar mass of urea (NH₂CO) = 60 g/mol
- Molar mass of KCl = 39 g/mol (K) + 35.5 g/mol (Cl) = 74.5 g/mol
- Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

### Step 3: Calculate the Depression in Freezing Point for Each Solute
The formula for depression in freezing point (ΔTf) is given by:
\[ \Delta T_f = K_f \cdot m \]
where \( K_f \) is the cryoscopic constant and \( m \) is the molality.

#### For Urea:
1. Calculate the number of moles of urea:
\[ \text{Moles of urea} = \frac{W_{urea}}{\text{Molar mass of urea}} = \frac{5 \text{ g}}{60 \text{ g/mol}} = 0.0833 \text{ mol} \]
2. Convert the mass of solvent to kg:
\[ \text{Mass of solvent} = 84 \text{ g} = 0.084 \text{ kg} \]
3. Calculate molality:
\[ m_{urea} = \frac{\text{Moles of urea}}{\text{Mass of solvent in kg}} = \frac{0.0833 \text{ mol}}{0.084 \text{ kg}} = 0.992 \text{ mol/kg} \]
4. Calculate ΔTf for urea:
\[ \Delta T_{f, urea} = K_f \cdot m_{urea} = 1.86 \text{ K kg/mol} \cdot 0.992 \text{ mol/kg} \approx 1.84 \text{ K} \]

#### For KCl:
1. Calculate the number of moles of KCl:
\[ \text{Moles of KCl} = \frac{W_{KCl}}{\text{Molar mass of KCl}} = \frac{1 \text{ g}}{74.5 \text{ g/mol}} \approx 0.0134 \text{ mol} \]
2. Calculate molality:
\[ m_{KCl} = \frac{0.0134 \text{ mol}}{0.084 \text{ kg}} \approx 0.1595 \text{ mol/kg} \]
3. Calculate ΔTf for KCl (considering it dissociates into K⁺ and Cl⁻, so i = 2):
\[ \Delta T_{f, KCl} = K_f \cdot (i \cdot m_{KCl}) = 1.86 \cdot (2 \cdot 0.1595) \approx 0.594 \text{ K} \]

#### For Glucose:
1. Calculate the number of moles of glucose:
\[ \text{Moles of glucose} = \frac{W_{glucose}}{\text{Molar mass of glucose}} = \frac{10 \text{ g}}{180 \text{ g/mol}} \approx 0.0556 \text{ mol} \]
2. Calculate molality:
\[ m_{glucose} = \frac{0.0556 \text{ mol}}{0.084 \text{ kg}} \approx 0.6619 \text{ mol/kg} \]
3. Calculate ΔTf for glucose (i = 1):
\[ \Delta T_{f, glucose} = K_f \cdot (i \cdot m_{glucose}) = 1.86 \cdot (1 \cdot 0.6619) \approx 1.23 \text{ K} \]

### Step 4: Calculate the Total Depression in Freezing Point
\[ \Delta T_f = \Delta T_{f, urea} + \Delta T_{f, KCl} + \Delta T_{f, glucose} \]
\[ \Delta T_f = 1.84 \text{ K} + 0.594 \text{ K} + 1.23 \text{ K} \approx 3.664 \text{ K} \]

### Step 5: Calculate the Freezing Point of the Solution
The freezing point of pure water is 273 K. Therefore, the freezing point of the solution is:
\[ T_f = T_{f, \text{pure}} - \Delta T_f = 273 \text{ K} - 3.664 \text{ K} \approx 269.36 \text{ K} \]

### Final Answer
The freezing point of the solution is approximately **269.63 K**.

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