Animated Solution for Chemistry - Solutions: If $\alpha$ is the degree of dissociation of ${\text{Na}}_2{\text{SO}}_4$, the van't Hoff factor ($i$) used for calculating the molecular mass is

Imagine dropping a pinch of sodium sulphate (${\text{Na}}_2{\text{SO}}_4$) into a beaker of water. It doesn't just sit there; it actively breaks apart! This process is called dissociation. But how do we quantify this breaking apart, especially when we want to calculate properties like boiling point elevation or freezing point depression? This is where the van't Hoff factor ($i$) steps in like a superhero.

The first thing we need to do is look at the chemical equation for the dissociation of sodium sulphate:

Look closely at the products. For every single molecule of ${\text{Na}}_2{\text{SO}}_4$ that breaks apart, we get two sodium ions (${\text{Na}}^{+}$) and one sulphate ion (${\text{SO}}_4^{2-}$).

If we add them up, the total number of ions produced per molecule, denoted by $n$, is:

Now, not every molecule might break apart. Sometimes, only a fraction of them do. This fraction is called the degree of dissociation ($\alpha$). To link the theoretical number of particles to the actual number of particles in the solution, we use the van't Hoff factor formula for dissociation:

This equation is a lifesaver in physical chemistry. It beautifully connects the microscopic world of ions ($n$) with the macroscopic observable behavior ($i$).

We already found that $n=3$. Let's carefully substitute this value into our master equation:

Simplifying the terms inside the bracket, $3-1$ gives us $2$. So, the expression becomes:

And there we have it! The van't Hoff factor for sodium sulphate, taking into account its degree of dissociation $\alpha$, is $1+2\alpha$. This perfectly matches option (b). It's a straightforward yet incredibly fundamental concept that frequently appears in competitive exams like JEE and NEET.