The Van’t Hoff factor $\left( {\rm{i}} \right)$for${K_2}S{O_4}$is….
A. $1$
B. $2$
C. $0$
D. $3$

Hint:
There will be two types of colligative properties one is observed colligative properties and one is normal colligative properties. And the ratio of both is known as Van’t Hoff factor.

Complete step by step solution
The Van’t Hoff factor is defined as the ratio of observed and normal colligative properties. In case of ionic compounds which are soluble in water Van’t Hoff factor provides the ratio of ions which forms after dissolving in water. For example in case of $N{a_2}S{O_4}$,this $N{a_2}S{O_4}$will breaks in water to form $2\,N{a^ + }$and one $SO_4^{2 - }$.So in this case our Van’t Hoff factor will be $2 + 1 = 3$.

In case of Osmotic pressure, Van’t Hoff factor is the ratio of observed osmotic effect and normal osmotic effect.
${\rm{i}} = \dfrac{{observed\,osmotic\,pressure}}{{normal\,osmotic\,pressure}}$

For the non-electrolytes, the value of Van’t Hoff Factor is $1$.

Whenever there occurs dissociation then the no of products must be greater than the number of reactants, then there exists a relationship between degree of dissociation and the Van’t Hoff factor which is given by
${\rm{i}} = \alpha n + \left( {1 - \alpha } \right) = 1 + \alpha \left( {n - 1} \right)$
where n is number of dissociated ions and $\alpha $ is the degree of dissociation.

So the Van't Hoff factor must be greater than 1.

Similarly, in case of association the number of reactants is greater than the number of products, so the relationship becomes,
\[{\rm{i}} = 1 - \left( {1 - \frac{1}{n}} \right)\alpha \]

So Van’t Hoff factor value must be less than `1

So, in case of ${K_2}S{O_4}$,there occurs $2{K^ + }$and $1\,SO_4^ - $
Hence, the Van't Hoff factor is 3.

Thus, the correct option is D.

Note:
Thus, the value of the Van't Hoff factor is the number of product particles by the number of reactant particles that is initially taken.