We define the Van't Hoff factor as the number of particles each solute formula unit breaks apart into when it dissolves.Here we are given $Na_2SO_4.10H_2O$The water of crystallization which $10H_2O$ will not be counted in the dissociation number.So, dissociation of a dilute solution of $Na2SO_4$ will be given as$Na2SO_4 ⇌2na^+ + SO_4^{2-}$In this chemical equation, we can see that there are $2Na^+$ and $1SO_4^{2-}$ ions. That makes the total number of ions equal to 3.So, that means Van’t Hoff factor for $Na_2SO_4.10H_2O$ is also 3.