An aqueous solution of urea has a freezing point of \(-0.52^oC\). Predict the osmotic pressure of the solution at \(37^oC\) [\(K_f\)  =1.86, assuming that the molar concentration and molality are numerically equal]

7.1 atm

The correct answer is option 2. 7.1 atm. 

Given,

Freezing Point Depression (\(\Delta T_f\))= \(-0.52^\circ\text{C}\)

Freezing Point Depression Constant (\(K_f\))= 1.86 °C kg/mol

Temperature for Osmotic Pressure Calculation (\(T\))= 37°C

Assumption: Molar concentration and molality are numerically equal.

The freezing point depression formula is:

\(\Delta T_f = K_f \cdot m\)

Where:

\(\Delta T_f\) = Freezing point depression = \(0.52^\circ\text{C}\) (positive value for calculation)

\(K_f\) = Freezing point depression constant = 1.86 °C kg/mol

\(m\) = Molality of the solution

Rearranging to solve for \(m\):

\(m = \frac{\Delta T_f}{K_f} = \frac{0.52}{1.86} \approx 0.28 \text{ mol/kg}\)

So, the molality \(m\) is approximately 0.28 mol/kg.

Assuming molality is numerically equal to molar concentration (since the density of the solution is close to that of water), the molar concentration of the solution is also 0.28 M.

The osmotic pressure (\(\Pi\)) of the solution is given by the formula:

\(\pi = M \cdot R \cdot T\)

Where:

\(M\) = Molar concentration = 0.28 M

\(R\) = Gas constant = 0.0821 L·atm/(mol·K)

\(T\) = Temperature in Kelvin = 37°C + 273.15 = 310.15 K

Substitute these values into the formula:

\(\pi = 0.28 \cdot 0.0821 \cdot 310.15 \approx 7.1 \text{ atm}\)

Summary: The osmotic pressure of the urea solution at 37°C is approximately 7.1 atm.

Therefore, the correct answer is: 2. 7.1 atm