Which one of the following will have the highest value of van’t Hoff factor, \(i\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(K_2SO_4\)
The van’t Hoff factor i is the number of particles one formula unit gives in solution. For strong electrolytes, which dissociate completely, i equals the number of ions formed:
\( \ce{K2SO4} \) → \( \ce{2K^{+}} \) + \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, so i = 3
\( \ce{MgSO4} \) → \( \ce{Mg^{2+}} \) + \( \ce{SO4^{2-}} \): 2 ions, so i = 2
KCl → \( \ce{K^{+}} \) + \( \ce{Cl^{-}} \): 2 ions, so i = 2
NaCl → \( \ce{Na^{+}} \) + \( \ce{Cl^{-}} \): 2 ions, so i = 2
So K2SO4 has the highest van’t Hoff factor.