Find the freezing point of a glucose solution whose osmotic pressure at `25^(@)C` is found to be 30 atm. `K_(f)` (water) `= 1.86 kg mol^(-1)K`.

Answer

To find the freezing point of a glucose solution whose osmotic pressure at 25°C is 30 atm, we can follow these steps:

### Step 1: Understand the relationship between osmotic pressure and molarity
We know that osmotic pressure (π) is given by the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \( \pi \) = osmotic pressure (in atm)
- \( i \) = van 't Hoff factor (which is 1 for glucose since it does not dissociate)
- \( C \) = molarity of the solution (in mol/L)
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin (25°C = 298 K)

### Step 2: Rearranging the formula to find molarity
Since \( i = 1 \) for glucose, we can rearrange the formula to find \( C \):
\[ C = \frac{\pi}{R \cdot T} \]

### Step 3: Substitute the values
Substituting the values into the equation:
\[ C = \frac{30 \, \text{atm}}{0.0821 \, \text{L·atm/(K·mol)} \cdot 298 \, \text{K}} \]

### Step 4: Calculate the molarity
Calculating the above expression:
\[ C = \frac{30}{0.0821 \cdot 298} = \frac{30}{24.4758} \approx 1.22 \, \text{mol/L} \]

### Step 5: Convert molarity to molality
Since we are using 1 kg of water as the solvent, the molality (m) can be approximated as equal to the molarity for dilute solutions:
\[ m \approx 1.22 \, \text{mol/kg} \]

### Step 6: Calculate the depression in freezing point
The depression in freezing point (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
For glucose, \( i = 1 \) and \( K_f \) for water is given as 1.86 °C kg/mol. Thus:
\[ \Delta T_f = 1 \cdot 1.86 \, \text{°C kg/mol} \cdot 1.22 \, \text{mol/kg} \]

### Step 7: Calculate the depression in freezing point
Calculating \( \Delta T_f \):
\[ \Delta T_f = 1.86 \cdot 1.22 \approx 2.27 \, \text{°C} \]

### Step 8: Determine the new freezing point
The normal freezing point of water is 0 °C. Therefore, the new freezing point of the solution is:
\[ \text{Freezing Point} = 0 \, \text{°C} - \Delta T_f = 0 - 2.27 \approx -2.27 \, \text{°C} \]

### Final Answer
The freezing point of the glucose solution is approximately **-2.27 °C**.

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