Decreasing order of freezing point of \(0.1\, \ M\) aqueous solution of following will be:

(A) \(K_4[Fe(CN)_6]\)

(B) \(K_2SO_4\)

(C) \(NH_2CONH_2\)

(D) \(AlCl_3\)

(E) \(HCl\)

Choose the correct answer from the options given below:

(C) > (E) > (B) > (D) > (A)

The correct answer is option 3. (C) > (E) > (B) > (D) > (A).

The freezing point depression of a solution depends on the number of solute particles present in the solution. The more particles there are, the greater the freezing point depression.

Here is the breakdown of each compound:

(C) \(NH_2CONH_2\) (Urea): Urea does not dissociate into ions when dissolved in water, so it only contributes one solute particle per molecule. Hence, its van't Hoff factor (\(i\)) is 1.

(E) \(HCl\): Hydrochloric acid dissociates completely in water into \(H^+\) ions and \(Cl^-\) ions. Therefore, it contributes two ions per molecule, resulting in a van't Hoff factor of 2.

(B) \(K_2SO_4\) (Potassium sulphate): When potassium sulphate dissolves in water, it dissociates into \(2K^+\) ions and \(SO_4^{2-}\) ions. So, it contributes three ions in total, giving it a van't Hoff factor of 3.

(D) \(AlCl_3\) (Aluminium chloride): Aluminium chloride dissociates into \(Al^{3+}\) ions and \(3Cl^-\) ions in solution. Thus, it contributes four ions per molecule, leading to a van't Hoff factor of 4.

(A) \(K_4[Fe(CN)_6]\) (Potassium ferrocyanide): This compound dissociates into \(4K^+\) ions and \(Fe(CN)_6^{4-}\) ions when dissolved in water. Consequently, it contributes a total of five ions, resulting in a van't Hoff factor of 5.

Given these van't Hoff factors and the fact that higher van't Hoff factors correspond to greater freezing point depression, the correct decreasing order of freezing points is: \(C\) > \(E\) > \(B\) > \(D\) > \(A\).

So, option 3: \(C\) > \(E\) > \(B\) > \(D\) > \(A\) is the correct answer.