Calculate the temperature at which a solution containing 54g of glucose `(C_6H_12O_6)`, in 250 g of water will freeze.

Answer

To calculate the temperature at which a solution containing 54 g of glucose (C₆H₁₂O₆) in 250 g of water will freeze, we will follow these steps:

### Step 1: Identify the given data
- Mass of glucose (solute), \( W_b = 54 \, \text{g} \)
- Mass of water (solvent), \( W_s = 250 \, \text{g} \)
- Freezing point depression constant for water, \( K_f = 1.86 \, \text{°C kg/mol} \)
- Molar mass of glucose, \( M = 180 \, \text{g/mol} \)

### Step 2: Convert the mass of the solvent from grams to kilograms
Since the freezing point depression constant \( K_f \) is given in °C kg/mol, we need to convert the mass of the solvent to kilograms:
\[ W_s = 250 \, \text{g} = \frac{250}{1000} = 0.250 \, \text{kg} \]

### Step 3: Calculate the number of moles of glucose
Using the molar mass of glucose, we can calculate the number of moles of glucose:
\[ \text{Number of moles of glucose} = \frac{W_b}{M} = \frac{54 \, \text{g}}{180 \, \text{g/mol}} = 0.3 \, \text{mol} \]

### Step 4: Calculate the freezing point depression (\( \Delta T_f \))
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \cdot \text{moles of solute} \cdot \frac{W_s}{1000} \]
Substituting the values:
\[ \Delta T_f = 1.86 \, \text{°C kg/mol} \cdot 0.3 \, \text{mol} \cdot 0.250 \, \text{kg} \]
Calculating:
\[ \Delta T_f = 1.86 \cdot 0.3 \cdot 0.250 = 0.1395 \, \text{°C} \]

### Step 5: Calculate the new freezing point of the solution
The normal freezing point of water is 0 °C. Therefore, the new freezing point (\( T_f \)) of the solution can be calculated as:
\[ T_f = 0 \, \text{°C} - \Delta T_f = 0 - 0.1395 = -0.1395 \, \text{°C} \]

### Step 6: Convert the freezing point to Kelvin
To convert the freezing point to Kelvin:
\[ T_f = -0.1395 + 273.15 = 272.99 \, \text{K} \]

### Final Answer
The temperature at which the solution will freeze is approximately **272.99 K**.

---