Which of the following aqueous solutions will have the lowest freezing point? Oa) 0.10 m sodium carbonate, Na₂CO3 Ob) 0.15 m calcium chloride, CaCl₂ O c) 0.20 methanol, CH3CH₂OH d) 0.30 m potassium chloride, KCI 0.25 m sodium nitrate, NaNO3
Which of the following aqueous solutions will have the lowest freezing point? Oa) 0.10 m sodium carbonate, Na₂CO3 Ob) 0.15 m calcium chloride, CaCl₂ O c) 0.20 methanol, CH3CH₂OH d) 0.30 m potassium chloride, KCI 0.25 m sodium nitrate, NaNO3
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:**Question:**
Which of the following aqueous solutions will have the lowest freezing point?
**Options:**
a) \(0.10 \, m\) sodium carbonate, \( \text{Na}_2\text{CO}_3\)
b) \(0.15 \, m\) calcium chloride, \( \text{CaCl}_2\)
c) \(0.20 \, m\) ethanol, \( \text{CH}_3\text{CH}_2\text{OH}\)
d) \(0.30 \, m\) potassium chloride, \( \text{KCl}\)
e) \(0.25 \, m\) sodium nitrate, \( \text{NaNO}_3\)
**Discussion:**
In terms of freezing point depression, the solution with the highest number of total ions will have the lowest freezing point. Consider the dissociation of each compound in water:
- Sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) dissociates to 3 ions: \(2 \, \text{Na}^+ + \text{CO}_3^{2-}\).
- Calcium chloride (\( \text{CaCl}_2 \)) dissociates to 3 ions: \( \text{Ca}^{2+} + 2 \, \text{Cl}^- \).
- Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \)) does not dissociate.
- Potassium chloride (\( \text{KCl} \)) dissociates to 2 ions: \( \text{K}^+ + \text{Cl}^- \).
- Sodium nitrate (\( \text{NaNO}_3 \)) dissociates to 2 ions: \( \text{Na}^+ + \text{NO}_3^- \).
Calculate the ion concentrations for each:
- a) \(0.10 \times 3 = 0.30 \, m\) ions
- b) \(0.15 \times 3 = 0.45 \, m\) ions
- c) Non-electrolyte, so \(0.20 \, m\) molecules
- d) \(0.30 \times 2 = 0.60 \, m\) ions
- e) \(0.25 \times 2 =
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