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Chapter 13: Problem 73

Arrange these solutions in order of decreasing freezing point: (a) \(0.10 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4},\) (b) \(0.35 \mathrm{~m} \mathrm{NaCl}\) (c) \(0.20 \mathrm{~m} \mathrm{MgCl}_{2},\) (d) \(0.15 \mathrm{~m} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) (e) \(0.15 \mathrm{~m}\) \(\mathrm{CH}_{3} \mathrm{COOH}\)

Short Answer

Expert verified
The order of solutions from highest freezing point to lowest: (d) C6H12O6, (e) CH3COOH, (a) Na3PO4, (c) MgCl2, (b) NaCl.

Step by step solution

01

Identify the number of solute particles

Identify how many particles each solute would dissociate into in solution: Na3PO4 dissociates into 4 particles (3 Na+ and 1 PO4-3), NaCl dissociates into 2 particles (Na+ and Cl-), MgCl2 dissociates into 3 particles (Mg2+ and 2 Cl-), C6H12O6 (glucose) and CH3COOH (acetic acid) do not ionize, so they form 1 particle each.
02

Calculate Van't Hoff factor for each solution

Calculate the Van't Hoff factor (i) for each solution multiplying the molality by the number of particles each compound forms: For (a) Na3PO4, i = 0.10 m * 4 = 0.4. For (b) NaCl, i = 0.35 m * 2 = 0.7. For (c) MgCl2, i = 0.20 m * 3 = 0.6. For (d) C6H12O6, i = 0.15 m * 1 = 0.15. For (e) CH3COOH, i = 0.15 m * 1 = 0.15.
03

Arrange the solutions

Arrange the solutions in order of decreasing i values, which represent decreasing freezing points: (b) NaCl > (c) MgCl2 > (a) Na3PO4 > (d) C6H12O6 = (e) CH3COOH.

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