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The van't Hoff factor $i$ for $0.1$ m NaCl (assuming complete dissociation) is:
A$1$, since solutions never increase in particle count
B$2$, since NaCl gives one Na$^+$ and one Cl$^-$ per unit
C$0.5$, since solubility is half for ionic salts here
D$3$, since NaCl forms three particles in solution
Answer & Solution
Correct answer: B. $2$, since NaCl gives one Na$^+$ and one Cl$^-$ per unit
NaCl $\to$ Na$^+$ + Cl$^-$: 2 particles, $i = 2$.
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