The depression in freezing point of 0.01 M aqueous solution of urea, sodium chloride and sodium sulphate is in the ratio:

1:2:3

The correct answer is option 2. 1:2:3.

The depression in the freezing point of a solution is given by the equation:

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

where \(\Delta T_f \) is the freezing point depression, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution.

For a dilute solution, the molality can be approximated as the molar concentration of the solute.

Given that the solutions are 0.01 M in concentration, we can assume that the molality is approximately 0.01 mol/kg.

Now, let's compare the depression in the freezing point for the three solutes:

1. Urea (\( CH_4N_2O \)):

The molecular formula of urea is \( CH_4N_2O \). It does not dissociate or ionize in water. Therefore, the number of particles it forms in the solution is 1. The depression in the freezing point for urea is given by:

\( \Delta T_{f_{urea}} = K_f \cdot m_{urea} \)

2. Sodium chloride (\( NaCl \)):

Sodium chloride dissociates into two ions (\( Na^+ \) and \( Cl^- \)) when dissolved in water. Therefore, the number of particles it forms in the solution is 2. The depression in the freezing point for sodium chloride is given by:

\( \Delta T_{f_{NaCl}} = 2 \cdot K_f \cdot m_{NaCl} \)

3. Sodium sulfate (\( Na_2SO_4 \)):

Sodium sulfate dissociates into three ions (2 \( Na^+ \) and \( SO_4^{2-} \)) when dissolved in water. Therefore, the number of particles it forms in the solution is 3. The depression in the freezing point for sodium sulfate is given by:

\(\Delta T_{f_{Na_2SO_4}} = 3 \cdot K_f \cdot m_{Na_2SO_4} \)

Comparing the ratios of the depressions at the freezing point, we have:

\(\Delta T_{f_{urea}} : \Delta T_{f_{NaCl}} : \Delta T_{f_{Na_2SO_4}} = 1 : 2 : 3 \)

Therefore, the depression in the freezing point ratio for the 0.01 M aqueous solutions of urea, sodium chloride, and sodium sulfate is (2) 1:2:3.