The van't Hoff Factor, \( i \) for complete dissociation of (i) \( NaCl \), (ii) \( MgSO_4 \) and (iii) \( K_2SO_4 \) are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (i) 2, (ii) 2, (iii) 3
The van’t Hoff factor i is the ratio of the number of particles actually present in solution to the number of formula units dissolved. For a strong electrolyte that dissociates completely, i is the number of ions given by one formula unit. Write the dissociation and count the ions:
\( \ce{NaCl -> Na+ + Cl-} \): 1 + 1 = 2 ions, so \( i = 2 \)
\( \ce{MgSO4 -> Mg^2+ + SO4^2-} \): 1 + 1 = 2 ions, so \( i = 2 \)
\( \ce{K2SO4 -> 2K+ + SO4^2-} \): 2 + 1 = 3 ions, so \( i = 3 \)
So the values are (i) 2, (ii) 2, (iii) 3. The charges on the ions do not matter: \( \ce{MgSO4} \) gives doubly charged ions but still only two particles, the same number as NaCl.