Chapter 13: Problem 69
Both \(\mathrm{NaCl}\) and \(\mathrm{CaCl}_{2}\) are used to melt ice on roads and sidewalks in winter. What advantages do these substances have over sucrose or urea in lowering the freezing point of water?
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Colligative Properties
- Freezing point depression
- Boiling point elevation
- Vapour pressure lowering
- Osmotic pressure
Van't Hoff Factor
- For example, \(\text{NaCl}\) splits into two particles: \( \text{Na}^+ \) and \( \text{Cl}^- \).
- \(\text{CaCl}_{2}\) splits into three particles: \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \) ions.
Ionic vs. Molecular Compounds
- Ionic Compounds: These consist of metals combined with non-metals and dissociate into ions in a solution. For instance, \(\text{NaCl}\) breaks into \( \text{Na}^+ \) and \( \text{Cl}^- \) ions, while \(\text{CaCl}_{2}\) dissociates into \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \) ions.
- Molecular Compounds: Composed of covalently bonded non-metal molecules, these do not break into ions in solution. A good example is sucrose, which stays intact as a whole molecule when dissolved in water.