Chapter 13: Problem 49
List the following aqueous solutions in order of increasing melting point. (The last three are all assumed to dissociate completely into ions in water. \()\) (a) \(0.1 \mathrm{m}\) sugar (b) \(0.1 \mathrm{m} \mathrm{NaCl}\) (c) \(0.08 \mathrm{m} \mathrm{CaCl}_{2}\) (d) \(0.04 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\)
Short Answer
Step by step solution
Understand the Concept of Melting Point Depression
Calculate Van't Hoff Factor for Each Solute
Calculate Effective Molality for Each Solution
List Solutions in Increasing Order of Melting Point
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Melting Point Depression
It doesn't depend on the type of solute you have. When you dissolve a non-volatile solute in a solvent, like salt in water, the crystals of the solvent require more energy to organize themselves.
Because of this disruption, the temperature at which the solid becomes a liquid is lowered. This is why substances like salt can melt ice on roads in winter.
The more solute particles there are, the more the melting point of the solution will drop below that of the pure solvent.
Van't Hoff Factor
This factor tells us how many particles a solute forms when it dissolves in a solution. It is particularly useful for calculating colligative properties.
- For substances like sugar (C鈧佲倐H鈧傗倐O鈧佲倎), which do not break apart into ions, the Van't Hoff factor is 1. - For ionic compounds like NaCl, which dissociate into ions, the factor increases. For instance, NaCl separates into two ions: Na鈦 and Cl鈦, so, i = 2.
- Some compounds create more particles, like CaCl鈧, disbanding into three ions: one Ca虏鈦 and two Cl鈦, giving i = 3. - Similarly, Na鈧係O鈧 becomes two Na鈦 and one SO鈧劼测伝, also resulting in i = 3.
This factor is crucial because it directly influences the extent to which the melting point will be depressed in solutions.
Aqueous Solutions
- Water is a versatile solvent due to its polarity, enabling it to dissolve a broad range of substances. - While dissolved, compounds can dissociate into ions, impacting properties like electrical conductivity and colligative effects.
For instance, when we dissolve NaCl in water, Na鈦 and Cl鈦 ions disperse throughout the solution. These ions can dramatically impact properties like boiling point, freezing point, and vapor pressure.
Colligative properties like melting point depression are especially prominent in aqueous solutions due to these dissolved particles.
Effective Molality
To calculate effective molality, you multiply the solution's molality by the Van't Hoff factor:
- Sugar: Since it doesn't dissociate, its effective molality is simply 0.1 molality 脳 1 (i = 1) = 0.1.
- NaCl: With an effective molality of 0.1 molality 脳 2 (i = 2) = 0.2, it has more particles affecting melting point depression.
- CaCl鈧: Its effective molality is higher at 0.08 molality 脳 3 (i = 3) = 0.24, reflecting more particles in the solution.
- Na鈧係O鈧: Effective molality here is 0.04 molality 脳 3 (i = 3) = 0.12.