The freezing point of of aqueous solution that contains `3%` urea. `7.45% KCl` and `9%` of glucose is (given `K_(f)` of water `=1.86` and asume molarity `=` molality)

Options

A. `290K`
B. `285.5K`
C. `267.42K`
D. `250K`

Explanation

To find the freezing point of the aqueous solution containing 3% urea, 7.45% KCl, and 9% glucose, we will follow these steps:

### Step 1: Assume the mass of the solvent
Assume the mass of water (solvent) to be 100 grams.

### Step 2: Calculate the mass of each solute
- **Urea**: 3% of 100 g = 3 g
- **KCl**: 7.45% of 100 g = 7.45 g
- **Glucose**: 9% of 100 g = 9 g

### Step 3: Calculate the number of moles of each solute
- **Molar mass of urea** = 60 g/mol
\[ \text{Moles of urea} = \frac{3 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ moles} \]

- **Molar mass of KCl** = 74.5 g/mol
\[ \text{Moles of KCl} = \frac{7.45 \text{ g}}{74.5 \text{ g/mol}} = 0.1 \text{ moles} \]

- **Molar mass of glucose** = 180 g/mol
\[ \text{Moles of glucose} = \frac{9 \text{ g}}{180 \text{ g/mol}} = 0.05 \text{ moles} \]

### Step 4: Calculate the total number of moles
\[ \text{Total moles} = \text{Moles of urea} + \text{Moles of KCl} + \text{Moles of glucose} = 0.05 + 0.1 + 0.05 = 0.2 \text{ moles} \]

### Step 5: Calculate the molality of the solution
Since we are assuming molarity equals molality:
\[ \text{Molality} = \frac{\text{Total moles}}{\text{Mass of solvent in kg}} = \frac{0.2 \text{ moles}}{0.1 \text{ kg}} = 2 \text{ mol/kg} \]

### Step 6: Calculate the change in freezing point (\( \Delta T_f \))
Using the formula:
\[ \Delta T_f = K_f \times m \]
Where \( K_f \) for water = 1.86 °C/m and \( m = 2 \text{ mol/kg} \):
\[ \Delta T_f = 1.86 \times 2 = 3.72 \text{ °C} \]

### Step 7: Calculate the freezing point of the solution
The freezing point of pure water is 0 °C. Therefore, the freezing point of the solution is:
\[ \text{Freezing point} = 0 - \Delta T_f = 0 - 3.72 = -3.72 \text{ °C} \]

### Final Answer
The freezing point of the aqueous solution is **-3.72 °C**.

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