The Van't Hoff factor for NaCl in ideal solution is approximately 2 because NaCl dissociates into how many particles?

The Van't Hoff factor for NaCl in ideal solution is approximately 2 because NaCl dissociates into how many particles?

• A. One
• B. Two
• C. Three
• D. Four

Answer: (B)

The Van't Hoff factor is a measure of how many particles a dissolved solute produces in solution. For NaCl, a strong electrolyte, it dissociates into two ions: Na+ and Cl−. That means each formula unit of NaCl yields two particles in solution, so the Van't Hoff factor is approximately 2 in an ideal solution. This is why colligative properties depend on particle count: doubling the number of dissolved particles doubles the effect. If a substance doesn’t dissociate, its i would be about 1 because it remains as a single particle; if more particles formed, i would be higher, but for NaCl under typical conditions it’s two.

The Van't Hoff factor is a measure of how many particles a dissolved solute produces in solution. For NaCl, a strong electrolyte, it dissociates into two ions: Na+ and Cl−. That means each formula unit of NaCl yields two particles in solution, so the Van't Hoff factor is approximately 2 in an ideal solution. This is why colligative properties depend on particle count: doubling the number of dissolved particles doubles the effect. If a substance doesn’t dissociate, its i would be about 1 because it remains as a single particle; if more particles formed, i would be higher, but for NaCl under typical conditions it’s two.