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 ($i$) for a solute that undergoes complete dissociation is equal to the total number of ions produced per formula unit of that solute.
Breakdown of Dissociation
(i) NaCl (Sodium Chloride):
Dissociates into one sodium ion and one chloride ion.
$NaCl \rightarrow Na^+ + Cl^-$
Total ions = $1 + 1 = 2$. Therefore, $i = 2$.
(ii) MgSO$_4$ (Magnesium Sulfate):
Dissociates into one magnesium ion and one sulfate ion.
$MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-}$
Total ions = $1 + 1 = 2$. Therefore, $i = 2$.
(iii) K$_2$SO$_4$ (Potassium Sulfate):
Dissociates into two potassium ions and one sulfate ion.
$K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$
Total ions = $2 + 1 = 3$. Therefore, $i = 3$.