The freezing point of an aqueous solution of non-electrolyte having an osmotic pressure of 2.0 atm at 300 K will be, `(K_(f) "for " H_(2)O = 1.86K mol^(-1)g)` and R = 0.0821 litre atm `K^(-1) mol^(-1)` . Assume molarity and molality to be same :

Options

A. `-0.151^(@)C`
B. `-1.151^(@)C`
C. `-3.151^(@)C`
D. `-2.151^(@)C`

Answer

To find the freezing point of an aqueous solution of a non-electrolyte with an osmotic pressure of 2.0 atm at 300 K, we can follow these steps:

### Step 1: Write the formula for osmotic pressure
The osmotic pressure (π) is given by the formula:
\[ \pi = iCRT \]
Where:
- \( \pi \) = osmotic pressure (atm)
- \( i \) = van 't Hoff factor (1 for non-electrolytes)
- \( C \) = molarity of the solution (mol/L)
- \( R \) = ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\))
- \( T \) = temperature (K)

### Step 2: Rearrange the formula to find molarity (C)
Since we know π, R, and T, we can rearrange the formula to find the molarity:
\[ C = \frac{\pi}{RT} \]

### Step 3: Substitute the known values
Substituting the values into the equation:
\[ C = \frac{2.0 \, \text{atm}}{0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}} \]

### Step 4: Calculate the molarity (C)
Calculating the value:
\[ C = \frac{2.0}{0.0821 \times 300} = \frac{2.0}{24.63} \approx 0.0812 \, \text{mol/L} \]

### Step 5: Calculate the depression in freezing point (ΔTf)
The depression in freezing point is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( K_f \) = freezing point depression constant (1.86 K kg/mol for water)
- \( m \) = molality (which we assume to be equal to molarity for this problem)

Since \( i = 1 \) for non-electrolytes:
\[ \Delta T_f = K_f \cdot C \]

### Step 6: Substitute the values to find ΔTf
Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.0812 \, \text{mol/L} \]
Calculating:
\[ \Delta T_f \approx 0.151 \, \text{K} \]

### Step 7: Determine the final freezing point (Tf)
The initial freezing point of pure water (Ti) is 0°C. The final freezing point (Tf) can be calculated as:
\[ T_f = T_i - \Delta T_f \]
Substituting the values:
\[ T_f = 0 - 0.151 = -0.151 \, \text{°C} \]

### Final Answer
The freezing point of the solution is approximately \(-0.151 \, \text{°C}\).

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