Question

The values of van’t Hoff factor for $\text{KCl}$, $\text{NaCl}$ and $K_{2}S{O}_4$ respectively are:
(A) $$1,1,2$$
(B) $$1,1,1$$
(C) $$2,2,3$$
(D) $$2,3,2$$

Answer 1

Hint: The van't Hoff factor is the quantitative measurement of a solute on the colligative property. The van’t Hoff factor is the number of particles into which the atom divides itself in the solution. It can also be defined as the number of ions formed by the compound in the solution, is known as van’t Hoff factor.

Complete Step by Step Solution:
The dissociation of $\text{KCl}$ is given by,
$$KCl\to K^{+}+C{l}^{-}$$
Therefore, van’t Hoff factor is 2 for $\text{KCl}$.
The dissociation of $\text{NaCl}$ is given by,
$NaCl\to N{a}^{+}+C{l}^{-}$
Therefore, van’t Hoff factor is 2 for$\text{NaCl}$.
The dissociation of $K_{2}S{O}_4$ is given by,
$$K_{2}S{O}_4\to 2K^{+}+S{O}_4^{2-}$$
Therefore, the van't Hoff factor is 3, as there are two ions of potassium and one sulphate ion.

Hence, (C) $$2,2,3$$ is the correct answer.

Additional information:
According to the question, the van’t Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. It is denoted by a small “i”. The value of van’t Hoff factor only concerns the number of ions, it doesn’t concern the charge of the ions. When a non-electrolytic substance is dissolved in water, the value of i is generally 1. However, when an ionic compound forms a solution in water, the value of i is equal to the total number of ions present in one formula unit of the substance.

Note: Clear concept of van’t Hoff factor and its properties. Colligative properties depend only on the number of solute particles, the dissociation of solute molecules into ions results in an increase in the number of particles and hence affect the colligative properties.