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:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. (C) > (E) > (B) > (D) > (A).
(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.
Summarize
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(A) ---- K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻ = 5 ions so i = 5
lowest freezing point
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(D) ----- AlCl₃ → Al³⁺ + 3Cl⁻ = 4 ions so i = 4
Low freezing point
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(B) ------- K₂SO₄ → 2K⁺ + SO₄²⁻ = 3 ions so i =3
Moderate freezing point
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(E) ----- HCl → H⁺ + Cl⁻ = 2 ions so i = 2
High freezing point
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(C) ------ NH₂CONH₂ = 1 ion so i =1
Highest freezing point
Freezing Point: Decreases as i increases. So the answer is option 3. (C) > (E) > (B) > (D) > (A).