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The van't Hoff factor (i) for sodium chloride (NaCl) in water is approximately:
A$1$ (no dissociation)
B$3$ (full dissociation into 3 ions)
C$0.5$ (association)
D$2$ (Na+ + Cl-)
Answer & Solution
Correct answer: D. $2$ (Na+ + Cl-)
NaCl dissociates almost completely in water into Na+ + Cl-, producing 2 particles per formula unit. So i ≈ 2 (theoretically 2; actual slightly less due to ion pairing). For non-electrolytes (glucose) i = 1. For CaCl2 (gives 3 ions) i = 3.
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