What is the van’t Hoff factor of the following:
a. Sucrose (C12H22O11)
b. Calcium Chloride (CaCl2)
c. Sodium Hydrogen Carbonate (NaHCO3)
d. Sodium Chloride (NaCl)
e. Copper (II) Sulfate (CuSO4)
The correct answer and explanation is:
Correct Answers:
a. Sucrose (C₁₂H₂₂O₁₁): 1
b. Calcium chloride (CaCl₂): 3
c. Sodium hydrogen carbonate (NaHCO₃): 2
d. Sodium chloride (NaCl): 2
e. Copper (II) sulfate (CuSO₄): 2
Explanation (300 words):
The van’t Hoff factor (i) represents the number of particles a solute produces in solution. It is used in colligative property calculations such as boiling point elevation, freezing point depression, and osmotic pressure. For non-electrolytes that do not dissociate in water, the van’t Hoff factor is 1. For electrolytes that dissociate into ions, the value corresponds to the total number of resulting particles.
a. Sucrose (C₁₂H₂₂O₁₁) is a covalent compound. It does not dissociate into ions when dissolved in water. Instead, it dissolves as intact molecules. Therefore, the van’t Hoff factor is 1.
b. Calcium chloride (CaCl₂) is an ionic compound. It dissociates completely in water into 1 Ca²⁺ ion and 2 Cl⁻ ions, giving a total of 3 particles. So, its van’t Hoff factor is 3.
c. Sodium hydrogen carbonate (NaHCO₃) dissociates into Na⁺ and HCO₃⁻ in solution, forming 2 ions. Thus, the van’t Hoff factor is 2.
d. Sodium chloride (NaCl) dissociates into Na⁺ and Cl⁻, giving 2 particles in solution. Therefore, the van’t Hoff factor is 2.
e. Copper (II) sulfate (CuSO₄) dissociates into Cu²⁺ and SO₄²⁻ in water. These are 2 distinct ions, so the van’t Hoff factor is 2. However, in certain conditions, copper sulfate can form complex ions or partially dissociate, but under normal dilute aqueous conditions, 2 is the accepted value.
These values assume complete dissociation for ionic compounds, which is generally valid for strong electrolytes in dilute solutions. For more concentrated solutions or weak electrolytes, the actual van’t Hoff factor may be lower due to ion pairing or incomplete dissociation.