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Consider the following aqueous solutions: (i) \(0.20 m \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (nonvolatile, nonelectrolyte); (ii) 0.10 \(m\) CaCl_i (iii) 0.12 \(m\) KBr; and (iv) \(0.12 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (a) Which solution has the highest boiling point? (b) Which solution has the lowest freezing point? (c) Which solution has the highest water vapor pressure?

Short Answer

Expert verified
(a) Solution (iv) has the highest boiling point. (b) Solution (iv) has the lowest freezing point. (c) Solution (i) has the highest vapor pressure.

Step by step solution

01

Understanding Colligative Properties

Colligative properties depend on the number of solute particles in solution, not their identity. These include boiling point elevation, freezing point depression, and vapor pressure lowering. We can compare these properties using the Van't Hoff factor (i) which indicates the number of particles the solute splits into.
02

Calculate Van't Hoff Factor for Each Solution

(i) 0.20 m HOCH鈧侰H鈧侽H (nonvolatile, nonelectrolyte): i = 1 (ii) 0.10 m CaCl鈧: i = 3 (CaCl鈧 dissociates into 3 ions: Ca虏鈦, 2 Cl鈦) (iii) 0.12 m KBr: i = 2 (KBr dissociates into 2 ions: K鈦, Br鈦) (iv) 0.12 m Na鈧係O鈧: i = 3 (Na鈧係O鈧 dissociates into 3 ions: 2 Na鈦, SO鈧劼测伝).
03

Determine Boiling Point Elevation

The boiling point elevation is given by \[ \Delta T_b = i \cdot K_b \cdot m \] where \(K_b\) is the ebullioscopic constant and \(m\) is molality. Since \(K_b\) is constant for all solutions, we only need to compare \(i \cdot m\):(i) 0.20 \(m\) \(\cdot\) 1 = 0.20 (ii) 0.10 \(m\) \(\cdot\) 3 = 0.30 (iii) 0.12 \(m\) \(\cdot\) 2 = 0.24 (iv) 0.12 \(m\) \(\cdot\) 3 = 0.36 Solution (iv) has the highest boiling point.
04

Determine Freezing Point Depression

The freezing point depression is given by \[ \Delta T_f = i \cdot K_f \cdot m \] where \(K_f\) is the cryoscopic constant. We compare \(i \cdot m\):(i) 0.20 \(m\) \(\cdot\) 1 = 0.20 (ii) 0.10 \(m\) \(\cdot\) 3 = 0.30 (iii) 0.12 \(m\) \(\cdot\) 2 = 0.24 (iv) 0.12 \(m\) \(\cdot\) 3 = 0.36 Solution (iv) has the lowest freezing point.
05

Determine Water Vapor Pressure

The vapor pressure lowering is related to the number of solute particles. A higher number of particles results in a lower vapor pressure.Compare \(i \cdot m\) again:(i) 0.20 \(m\) \(\cdot\) 1 = 0.20 (ii) 0.10 \(m\) \(\cdot\) 3 = 0.30 (iii) 0.12 \(m\) \(\cdot\) 2 = 0.24 (iv) 0.12 \(m\) \(\cdot\) 3 = 0.36 Solution (i) has the highest vapor pressure.

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Key Concepts

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

Boiling Point Elevation
Boiling point elevation refers to the phenomenon where the boiling point of a solvent increases when a solute is dissolved in it. This occurs because the addition of solute particles disrupts the solvent's ability to vaporize, requiring more heat to reach the boiling point. This is quantified by the formula: \[ \Delta T_b = i \cdot K_b \cdot m \]
  • \( \Delta T_b \) is the change in boiling point.
  • \( i \) is the Van't Hoff factor, representing the number of particles the solute forms in solution.
  • \( K_b \) is the ebullioscopic constant, a property of the solvent.
  • \( m \) is the molality of the solution.
For example, in a solution like Na鈧係O鈧, which separates into three ions (2 Na鈦 and SO鈧劼测伝), the Van't Hoff factor is greater, leading to a higher boiling point when compared to solutions with solutes that dissociate into fewer particles or none at all.
Freezing Point Depression
Freezing point depression is observed when the freezing point of a solvent decreases upon addition of a solute. The solute particles disrupt the formation of a solid structure of the solvent, thus requiring a lower temperature to freeze. The formula used here is similar: \[ \Delta T_f = i \cdot K_f \cdot m \]
  • \( \Delta T_f \) represents the change in freezing point.
  • \( i \) is the Van't Hoff factor.
  • \( K_f \) is the cryoscopic constant of the solvent.
  • \( m \) represents the solution's molality.
The more particles the solute dissociates into, the more significant the freezing point depression. Therefore, a solution like Na鈧係O鈧 with a Van't Hoff factor of 3 will have a more pronounced lowering of the freezing point than a solution like KBr, which dissociates into two particles.
Van't Hoff Factor
The Van't Hoff factor, denoted as \( i \), is crucial in understanding colligative properties. It signifies the number of particles generated from the dissolution of one mole of solute in a solution. For nonelectrolytes, which don't dissociate into ions, \( i \) is typically 1. For electrolytes like CaCl鈧, which dissociates into three ions (Ca虏鈦 and 2 Cl鈦), \( i \) is 3. With higher \( i \) values, colligative properties such as boiling and freezing points are more affected. This is because more particles lead to more significant disruption of the solvent鈥檚 properties. In sum, the Van't Hoff factor directly informs us about the extent to which colligative properties will change.
Vapor Pressure Lowering
Vapor pressure lowering is another important colligative property. It occurs when the addition of a nonvolatile solute reduces the solvent's vapor pressure. The reduced pressure results because solute particles occupy space at the liquid's surface, hindering solvent molecules from escaping into the gas phase. The principle is straightforward: more solute particles result in greater vapor pressure lowering due to increased interference at the surface. Thus, a solution with a lower Van't Hoff factor, meaning fewer dissociated particles such as HOCH鈧侰H鈧侽H with \( i = 1 \), will have a higher vapor pressure compared to solutions with electrolytes that break into more particles.

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Most popular questions from this chapter

Concentrated sulfuric acid has a density of \(1.84 \mathrm{g} /\) \(\mathrm{cm}^{3}\) and is \(95.0 \%\) by weight \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) What is the molality of this acid? What is its molarity?

Arrange the following aqueous solutions in order of (i) increasing vapor pressure of water and (ii) increasing boiling point. (a) \(0.35 \mathrm{m} \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (a nonvolatile solute) (b) \(0.50 \mathrm{m}\) sugar (c) \(0.20 \mathrm{m} \mathrm{KBr}\) (a strong electrolyte) (d) \(0.20 m \mathrm{Na}_{2} \mathrm{SO}_{4}\) (a strong electrolyte)

Pure iodine \((105 \mathrm{g})\) is dissolved in \(325 \mathrm{g}\) of \(\mathrm{CCl}_{4}\) at \(65^{\circ} \mathrm{C} .\) Given that the vapor pressure of \(\mathrm{CCl}_{4}\) at this temperature is \(531 \mathrm{mm}\) Hg, what is the vapor pressure of the \(\mathrm{CCl}_{4}-\mathrm{I}_{2}\) solution at \(65^{\circ} \mathrm{C} ?\) (Assume that \(\mathrm{I}_{2}\) does not contribute to the vapor pressure.

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