The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximately:
- $1$
- $0.5$
- $2$
- $3$
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Correct answer: C. $2$
NaCl dissociates in water: $\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$. One formula unit produces two particles in solution.
Van't Hoff factor $i = \dfrac{\text{actual moles of particles}}{\text{moles of solute formula units}}$. For complete dissociation of NaCl, $i = 2$.
In reality, ion pairing slightly reduces $i$ below the theoretical value, especially at higher concentrations.
At dilute $0.01$ M the assumption of complete dissociation is reasonable, and the experimental $i$ is close to $1.9$.
The question asks for the theoretical value assuming complete dissociation, so 2 is the cleanest answer.
Compare with $\mathrm{CaCl_2}$: dissociates into 3 ions, so $i = 3$ (option D) — common trap.