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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

Expert verified
Increasing order of melting point: CaCl鈧, NaCl, Na鈧係O鈧, Sugar.

Step by step solution

01

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.
02

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鈦 and Cl鈦, CaCl鈧 into Ca虏鈦 and two Cl鈦, and Na鈧係O鈧 into two Na鈦 and SO鈧劼测伝.
03

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: \(0.04 \, m imes 3 = 0.12\).
04

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. Na鈧係O鈧 (\(0.12 \, m_{eff}\))3. NaCl (\(0.2 \, m_{eff}\))4. CaCl鈧 (\(0.24 \, m_{eff}\)).
05

List in Increasing Melting Point Order

The order of increasing melting point (decreasing depression of melting point) is: 1. CaCl鈧 2. NaCl 3. Na鈧係O鈧 4. Sugar. The more the solution depresses the melting point, the lower the melting point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \(i = 2\).
  • **CaCl鈧 (Calcium Chloride):** Breaks into three ions: \(\text{Ca}^{2+}\) and two \(\text{Cl}^-\), resulting in \(i = 3\).
  • **Na鈧係O鈧 (Sodium Sulfate):** Dissociates into three ions: two \(\text{Na}^+\) and one \(\text{SO}_4^{2-}\), hence \(i = 3\).
Understanding the Van鈥檛 Hoff factor is key for predicting how a solute will affect the solution鈥檚 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鈥檛 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鈥檛 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:** \(m_{\text{eff}} = 0.1 \, m \times 2 = 0.2\)
  • **CaCl鈧:** \(m_{\text{eff}} = 0.08 \, m \times 3 = 0.24\)
  • **Na鈧係O鈧:** \(m_{\text{eff}} = 0.04 \, m \times 3 = 0.12\)
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, CaCl鈧, and Na鈧係O鈧. 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.

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Most popular questions from this chapter

A \(10.7 \mathrm{m}\) solution of \(\mathrm{NaOH}\) has a density of \(1.33 \mathrm{g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). Calculate the following: (a) the mole fraction of \(\mathrm{NaOH}\) (b) the weight percent of \(\mathrm{NaOH}\) (c) the molarity of the solution

You wish to prepare an aqueous solution of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3},\) in which the mole fraction of the solute is \(0.093 .\) What mass of glycerol must you add to \(425 \mathrm{g}\) of water to make this solution? What is the molality of the solution?

Starch contains \(\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{O},\) and \(\mathrm{O}-\mathrm{H}\) bonds. Hydrocarbons have only \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{H}\) bonds. Both starch and hydrocarbons can form colloidal dispersions in water. Which dispersion is classified as hydrophobic? Which is hydrophilic? Explain briefly.

Which substance would have the greater influence on the vapor pressure of water when added to \(1000 .\) g of the liquid: \(10.0 \mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) or \(10.0 \mathrm{g}\) of ethylene glycol \(\left[\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right]\) ?

Assume you dissolve \(45.0 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O},\) in \(425 \mathrm{mL}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is \(0.785 \mathrm{g} / \mathrm{mL}\).)

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