Question

Van’t Hoff’s factor of aq. ${K_2}S{O_4}$ at infinite dilution has value equal to:
A.1
B.2
C.3
D.Between 2 and 3

Answer 1

We need to know that the Van’t Hoff factor is really the ratio between the particular concentration of particles produced upon dissolution and therefore the concentration of substance before dilution. For non-electrolysis, the Van’t Hoff factor is unity (1).

Complete answer:
We have to know that the Van’t Hoff factor may be a measure of the effect of a solute on colligative properties like osmotic pressure, boiling-point elevation and relative lowering in vapour pressure and freezing-point depression.
It is denoted by ‘i’ it is actually the ratio between the actual concentration of particles produced upon dilution and the concentration of substance before dilution. For most ionic compounds, once they are dissolved in water, the Van’t Hoff factor is adequate to the entire number of discrete ions produced by the substance.
For most non-ionic compounds, the Van’t Hoff factor is 1 as no other ion is produced by them upon dilution. At any time, a small number of ions pair with one another and are counted as a single particle. Due to this reason, the Van’t Hoff factor is less than what predicted in an ideal solution.
In the reaction, ${K_2}S{O_4} \to 2{K^ + } + SO_4^{2 - }$
At infinite dilution, each molecule gives 3 ions, hence the Van’t Hoff factor, i = 3.
Therefore, the solution to this present question is option C that is 3.

Note:
We have to remember that the Van’t Hoff factor for a substance is often less but one also. This is possible when the ions of a substance dissolved in a solution come together or associate with each other.