van't hoff factor

Question

Which solute has van't Hoff factor i ≈ 3 in dilute aqueous solution?

RankGuru · Accepted Answer

1. **The Concept**: $i = \frac{	ext{Actual number of particles after dissociation}}{	ext{Theoretical number of particles}}$.
2. **Evaluation**:
   - **Glucose**: Non-electrolyte. $i = 1$.
   - **$KCl$**: Dissociates into $K^+$ and $Cl^-$. $i = 2$.
   - **$Na_2SO_4$**: Dissociates into $2 Na^+$ and $1 SO_4^{2-}$. Total particles = $3$. Thus, **$i = 3$**.
   - **$AlCl_3$**: Dissociates into $1 Al^{3+}$ and $3 Cl^-$. Total particles = $4$. Thus, $i = 4$.
3. **Prerequisite**: Complete dissociation is assumed for strong electrolytes in very dilute solutions.

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van't hoff factor

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More van't hoff factor Practice Questions

The Van't Hoff factor (i) for NaCl in dilute aqueous solution is approximately: