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: Convert the given temperature to Kelvin
The temperature given is 25°C. To convert this to Kelvin:
\[ T(K) = T(°C) + 273.15 = 25 + 273.15 = 298.15 \, K \]

### Step 2: Use the osmotic pressure formula to find the concentration
The formula for osmotic pressure (\(\Pi\)) is given by:
\[ \Pi = C \cdot R \cdot T \]
Where:
- \(\Pi\) = osmotic pressure (30 atm)
- \(C\) = concentration in molarity (mol/L)
- \(R\) = universal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (298.15 K)

Rearranging the formula to find \(C\):
\[ C = \frac{\Pi}{R \cdot T} = \frac{30 \, \text{atm}}{0.0821 \, \text{L·atm/(K·mol)} \cdot 298.15 \, K} \]
Calculating \(C\):
\[ C = \frac{30}{0.0821 \times 298.15} \approx 1.23 \, \text{mol/L} \]

### Step 3: Assume molarity equals molality
Assuming the density of the solution is approximately 1 g/mL, we can approximate molarity (mol/L) to molality (mol/kg):
\[ \text{molality} \approx 1.23 \, \text{mol/kg} \]

### Step 4: Calculate the change in freezing point (\(\Delta T_f\))
The change in freezing point is given by the formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \(K_f\) = freezing point depression constant for water (1.86 °C·kg/mol)
- \(m\) = molality (1.23 mol/kg)

Calculating \(\Delta T_f\):
\[ \Delta T_f = 1.86 \, °C \cdot 1.23 \, \text{mol/kg} \approx 2.29 \, °C \]

### Step 5: Calculate the freezing point of the solution
The freezing point of the solution (\(T_f\)) can be found using:
\[ T_f = T_f^0 - \Delta T_f \]
Where:
- \(T_f^0\) = freezing point of pure water (0 °C)

Calculating \(T_f\):
\[ T_f = 0 - 2.29 \approx -2.29 \, °C \]

### Final Answer
The freezing point of the glucose solution is approximately:
\[ \boxed{-2.29 \, °C} \]