Problem 49 - Chemistry and Chemical Reactivity

Question

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) \(0.04 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Step-by-Step Solution

Answer

Order: (a) Sugar, (d) Na₂SO₄, (b) NaCl, (c) CaCl₂.

Step 1: Understand the Concept of Melting Point Depression

Melting point depression is a colligative property, which means it depends on the number of solute particles in a solution, not the identity of the solute. When a non-volatile solute is added to a solvent, the melting point of the solution is lower than that of the pure solvent.

Step 2: Calculate Van't Hoff Factor for Each Solute

For ideal solutions, calculate the Van't Hoff factor (i) which is the number of particles the solute dissociates into: - Sugar (C₁₂H₂₂O₁₁) does not dissociate, so for sugar, i = 1. - NaCl dissociates into 2 ions (Na⁺, Cl⁻), so i = 2. - CaCl₂ dissociates into 3 ions (Ca²⁺, 2Cl⁻), so i = 3. - Na₂SO₄ dissociates into 3 ions (2Na⁺, SO₄²⁻), so i = 3.

Step 3: Calculate Effective Molality for Each Solution

The effective molality is calculated by multiplying the given molality by the Van't Hoff factor. - Sugar: effective molality = 0.1 × 1 = 0.1. - NaCl: effective molality = 0.1 × 2 = 0.2. - CaCl₂: effective molality = 0.08 × 3 = 0.24. - Na₂SO₄: effective molality = 0.04 × 3 = 0.12.

Step 4: List Solutions in Increasing Order of Melting Point

The solution with the smallest effective molality will have the highest melting point. Order the solutions as follows: 1. Sugar (0.1m, i=1) - effective molality 0.1 2. Na₂SO₄ (0.04m, i=3) - effective molality 0.12 3. NaCl (0.1m, i=2) - effective molality 0.2 4. CaCl₂ (0.08m, i=3) - effective molality 0.24.

Key Concepts

Melting Point Depression

Melting point depression is a fascinating concept in chemistry. It is a type of colligative property, which means it relies only on the number of solute particles in a solution.
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

The Van't Hoff factor, represented as "i," is critical in understanding how solutes affect solution properties.
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 (C₁₂H₂₂O₁₁), 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⁺ and Cl⁻, so, i = 2.
- Some compounds create more particles, like CaCl₂, disbanding into three ions: one Ca²⁺ and two Cl⁻, giving i = 3. - Similarly, Na₂SO₄ becomes two Na⁺ and one SO₄²⁻, 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

An aqueous solution has water as the solvent, which plays a big role in everyday life and chemistry. Many reactions occur in aqueous solutions, where substances are dissolved in water.
- 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⁺ and Cl⁻ 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

Effective molality is an important concept when examining colligative properties. It's not just about the amount of solute but also considers how many particles that solute breaks into in the solution.
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 × 1 (i = 1) = 0.1.
• NaCl: With an effective molality of 0.1 molality × 2 (i = 2) = 0.2, it has more particles affecting melting point depression.
• CaCl₂: Its effective molality is higher at 0.08 molality × 3 (i = 3) = 0.24, reflecting more particles in the solution.
• Na₂SO₄: Effective molality here is 0.04 molality × 3 (i = 3) = 0.12.

By understanding effective molality, you can predict how much the melting point will decrease in a given solution, which is invaluable in organizing solutions and comparing their properties.