The molecular weights determined by using colligative properties of substances which associate or dissociate will be abnormal and are called abnormal molecular weights. Ionic substances like NaCl, BaCl₂, AlCl₃, etc., ionize in solutions. Their colligative properties are high due to increase in number of particles when compared with solutions of non-electrolytes like glucose having equal molecular mass. When molecules of substance like acetic acid in benzene associate as dimers, trimers or polymers, the number of particles decreases and their colligative properties also decrease. The ratio of the observed colligative property and calculated colligative property is called Van’t Hoff factor i.

\[\text{i = }\frac{\text{observed colligative property}}{\text{calculated colligative property}}\]

\[\text{or, i = }\frac{\text{calculated molecular mass}}{\text{observed molecular mass}}\]

The value of Van’t Hoff factor i is greater than 1 for ionic substances while it has lower value than 1 for associated substances.

\[\frac{\text{calculated molecular mass}}{\text{observed molecular mass}} = \frac{\text{Normal number of solute particles}}{\text{Number of solute particles after dissociation or association}}\]

For which of the following in dilute solution, Van’t Hoff factor i will be equal to 3?

Na₂SO₄

The correct answer is option 4. \(Na_2SO_4\).

(1) NaCl (sodium chloride) is an ionic compound and a strong electrolyte. In aqueous solution, NaCl completely dissociates into \(Na^+\) and \(Cl^-\) ions. So, the Van't Hoff factor for NaCl is \(i = 2\), not 3.

(2) Urea is a non-electrolyte and does not dissociate into ions in solution. Therefore, the Van't Hoff factor for urea is \(i = 1\)

(3) \(MgSO_4\) (magnesium sulfate) is an ionic compound and a strong electrolyte. In aqueous solution, \(MgSO_4\) dissociates into \(Mg^{2+}\) and \(SO_4^{2-}\) ions. So, the Van't Hoff factor for \(MgSO_4\) is \(i = 2\), not 3.

(4) \(Na_2SO_4\) (sodium sulfate) is an ionic compound and a strong electrolyte. In aqueous solution, \(Na_2SO_4\) dissociates into 2 \(Na^+\) ions and 1 \(SO_4^{2-}\) ion. So, the Van't Hoff factor for \(Na_2SO_4\) is \(i = 3\).

Therefore, among the given substances, option (4) \(Na_2SO_4\) is the one for which the Van't Hoff factor (\(i\)) will be equal to 3 in a dilute solution.