Question

The Van’t Hoff factor (i) for ${ Na }_{ 2 }{ SO }_{ 4 }$ is:
(A) 3
(B) 2
(C) 1
(D) 0

Answer 1

Hint : To answer this question you should know that for salt the Van’t Hoff factor is equal to the total number of ions that it dissociates into, in a solvent. Now try to find the answer to this question accordingly.

Complete step by step solution :The van 't Hoff factor (i) is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering in vapor pressure, boiling-point elevation, and freezing-point depression.
The Van't Hoff factor is always positive. It can’t be zero.

We define the Van't Hoff factor (i) as the number of particles each solute formula unit breaks apart into when it dissolves.
Here we are given ${ Na }_{ 2 }{ SO }_{ 4 }$
So, dissociation of a dilute solution of ${ Na }_{ 2 }{ SO }_{ 4 }$ will be given as
$Na_{ 2 }SO_{ 4 }\quad \rightleftharpoons \quad 2Na^{ + }+SO_{ 4 }^{ 2- }$

In this chemical equation, we can see that there are 2 $Na^{ + }$ and 1 $SO_{ 4 }^{ 2- }$ ions. That makes the total number of ions equal to 3.
So, that means Van’t Hoff factor for (i) for ${ Na }_{ 2 }{ SO }_{ 4 }$ is also 3.
Therefore, we can conclude that the correct answer to this question is option A.

Note : We should also know that the typical definition is that it is the ratio of the actual concentration of particles in solution to the concentration of the solute as calculated by its mass. So for non-ionic compounds in solution, like glucose (${ C }_{ 6 }{ H }_{ 12 }{ O }_{ 6 }$), the van't Hoff factor is 1.