If a is the degree of dissociation of \(Na_2SO_4\) the van’t Hoff factor \((i)\) used for calculating the molecular mass is:

\(1 + 2\alpha \)

The correct answer is option 3. \(1 + 2\alpha \)

The balanced chemical equation for the ionization of sodium sulphate (\(Na_2SO_4\)) is:

\(Na_2SO_4 \rightarrow 2 Na^+ + SO_4^{2-}\)

In this equation, one molecule of sodium sulphate dissociates to produce two sodium cations (\(Na^+\)) and one sulphate anion (\(SO_4^{2-}\)).

Suppose we have \(n\) molecules of sodium sulphate, out of which \(n\alpha\) molecules undergo dissociation, where \(\alpha\) is the degree of dissociation. Therefore, the dissociation of \(n\alpha\) molecules of sodium sulphate will result in the formation of \(2n\alpha\) sodium cations and \(n\alpha\) sulphate anions.

The remaining undissociated molecules of sodium sulphate will be \(n(1-\alpha)\).

To determine the total number of particles (ions) produced by the dissociation of \(n\) molecules of sodium sulphate, we add the number of sodium ions (\(2n\alpha\)) and the number of sulphate ions (\(n\alpha\)), which gives us a total of \(3n\alpha\) ions.

To the above number, add the number of undissociated sodium sulphate molecules to obtain the total number of particles present after dissociation

\(n(1−\alpha ) + 3n\alpha = n(1 + 2\alpha)\)

To calculate the van't Hoff factor (\(i\)) used for calculating the molecular mass, we divide the total number of particles (ions) by the total number of sodium sulphate molecules:

\(i = \frac{\text{Total number of particles present after dissociation}}{\text{Total number of sodium sulphate molecules}} = \frac{{ n(1 + 2\alpha)}}{{n}} = 1 + 2\alpha\)

Therefore, the van't Hoff factor (\(i\)) used for calculating the molecular mass of sodium sulphate is \(1 + 2\alpha \).