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

Increasing order of melting point: CaCl₂, NaCl, Na₂SO₄, Sugar.

Step 1: Understand Colligative Properties

The melting point of a solution is influenced by the number of dissolved solute particles. The presence of solute particles lowers the freezing point of a solvent, a colligative property. The greater the number of particles, the lower the freezing (or melting) point.

Step 2: Calculate Van't Hoff Factor

The Van't Hoff factor (\(i\)) indicates the number of particles that a solute forms in solution. For sugar (a non-electrolyte), \(i = 1\). For electrolytes, \(i\) corresponds to the number of ions the compound dissociates into: \(i = 2\) for NaCl, \(i = 3\) for CaCl_2, and \(i = 3\) for Na_2SO_4 because NaCl dissociates into Na⁺ and Cl⁻, CaCl₂ into Ca²⁺ and two Cl⁻, and Na₂SO₄ into two Na⁺ and SO₄²⁻.

Step 3: Calculate Effective Molality

Effective molality is the product of the molality of the solution and the Van't Hoff factor: \(m_{eff} = m imes i\). For each solution, calculate the effective molality:- \(0.1 \, m\) sugar: \(0.1 \, m imes 1 = 0.1\)- \(0.1 \, m\) NaCl: \(0.1 \, m imes 2 = 0.2\)- \(0.08 \, m\) CaCl_2: \(0.08 \, m imes 3 = 0.24\)- \(0.04 \, m\) Na_2SO_4: \(0.04 \, m imes 3 = 0.12\).

Step 4: Order Solutions by Effective Molality

The solutions with higher effective molality will have lower melting points. Therefore, order the solutions by increasing effective molality:1. Sugar (\(0.1 \, m_{eff}\))2. Na₂SO₄ (\(0.12 \, m_{eff}\))3. NaCl (\(0.2 \, m_{eff}\))4. CaCl₂ (\(0.24 \, m_{eff}\)).

Step 5: List in Increasing Melting Point Order

The order of increasing melting point (decreasing depression of melting point) is: 1. CaCl₂ 2. NaCl 3. Na₂SO₄ 4. Sugar. The more the solution depresses the melting point, the lower the melting point.

Key Concepts

Van't Hoff factor

The Van't Hoff factor, symbolized as \(i\), plays a crucial role in understanding the behavior of solutions, especially when it comes to colligative properties. It indicates the number of particles a solute yields after dissolving in a solution. For non-electrolytes, like sugar, \(i = 1\) because they dissolve without breaking into ions.
Electrolytes, on the other hand, dissociate into ions, resulting in an \(i\) value greater than 1. For example:

• **NaCl (Sodium Chloride):** Dissociates into two ions: \(\text{Na}^+\) and \(\text{Cl}^-\), giving \(i = 2\).
• **CaCl₂ (Calcium Chloride):** Breaks into three ions: \(\text{Ca}^{2+}\) and two \(\text{Cl}^-\), resulting in \(i = 3\).
• **Na₂SO₄ (Sodium Sulfate):** Dissociates into three ions: two \(\text{Na}^+\) and one \(\text{SO}_4^{2-}\), hence \(i = 3\).

Understanding the Van’t Hoff factor is key for predicting how a solute will affect the solution’s freezing point and other colligative properties. This relation forms the basis for calculating effective molality.

Effective molality

Effective molality is a concept used to quantify the real impact of solutes on colligative properties, like freezing point depression. It combines a solute's molality with its Van’t Hoff factor to give a more accurate picture of a solution's behavior.
The formula for effective molality is:
\[m_{\text{eff}} = m \times i\]where \(m\) is the molality of the solute and \(i\) is the Van’t Hoff factor.
Let's see how this applies to our previous examples:

• **Sugar (Non-electrolyte):** \(m_{\text{eff}} = 0.1 \, m \times 1 = 0.1\)
• **NaCl:** \(m_{\text{eff}} = 0.1 \, m \times 2 = 0.2\)
• **CaCl₂:** \(m_{\text{eff}} = 0.08 \, m \times 3 = 0.24\)
• **Na₂SO₄:** \(m_{\text{eff}} = 0.04 \, m \times 3 = 0.12\)

By determining effective molality, we can better compare the colligative effects of different solutes in solutions. Solutions with higher effective molality impact the freezing point more substantially.

Electrolytes

Electrolytes are substances that, when dissolved in water, dissociate into ions and allow the solution to conduct electricity. This dissociation affects colligative properties like freezing point depression. Electrolytes are classified into two broad types:

• **Strong Electrolytes:** Completely dissociate into ions in solution. Examples include NaCl, CaCl₂, and Na₂SO₄. These substances contribute to a higher effective number of particles, influencing colligative properties significantly.
• **Weak Electrolytes:** Partially dissociate in solution, leading to fewer ions. They have a less pronounced effect on colligative properties, which is often accounted for by a smaller Van't Hoff factor.

In the context of colligative properties, it is essential to consider whether a solute is an electrolyte and, if so, its strength, to evaluate its impact comprehensively. This knowledge is particularly useful for predicting and explaining phenomena such as freezing point depression, where electrolytes play a crucial role.

Freezing Point Depression

Freezing point depression is a colligative property that describes the lowering of a solvent's freezing point due to the addition of solute. This property is a direct result of the number of solute particles present in the solution.
Freezing point depression can be calculated using the formula:
\[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression, \(i\) is the Van't Hoff factor, \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.
It's important to consider:

• **Higher effective molality:** Leads to greater freezing point depression since more particles interfere with the formation of a solid structure.
• **Type of Solute:** Non-electrolytes and electrolytes behave differently due to the number of particles they produce.

Knowing the freezing point depression allows for the ordering of solutions by how much they lower the solvent's freezing point. This understanding is crucial in applications like de-icing roads, where substances like NaCl lower the freezing point of water effectively.