Question

The Van't Hoff factor of NaCl assuming 100% dissociation is :
(A) 1/2
(B) 2
(C) 1
(D) 3

Answer 1

Hint: Van't Hoff factor is taken into consideration when the solute is ionizable when added to the solvent i.e. the solute is an ionic compound. For non-ionic solutes the Van't Hoff factor is taken as 1 as the number of particles of solute do not increase upon addition to the solvent to form a solution.

Complete step by step answer:
The Van't Hoff factor ( symbol is "i") is a measure of the effect of a solute on colligative properties of the solution.
Colligative properties are the properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.
The number ratio can be related to the various units for concentration of a solution, for example, molarity, molality, normality etc.
The colligative properties are:
Relative lowering of vapor pressure
Elevation of boiling point
Depression of freezing point
Osmotic pressure

The Van't Hoff factor is the ratio between the actual concentration of particles produced after the solute is dissolved and the concentration of particles as calculated from its molecular mass.
In general, the Van't Hoff factor for ionic compounds is the number of individual ions present in the compound.
Based on the above method we will calculate Van't Hoff factor for NaCl.
Sodium chloride consists of two ions i.e. the sodium ion and the chlorine ion. So, the Van't Hoff factor for NaCl considering complete dissociation is 2.
So, the correct answer is “Option B”.

Note: In the above section we saw that the Van't Hoff factor is greater than 1 in case of dissociation. However, the Van't Hoff factor is less than 1 when the solute particles associate in the solvent. For example, the Van't Hoff factor for benzoic acid is 0.5 in benzene as the acid undergoes association.