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Chapter 12: Problem 14

Explain why the boiling point of a solution containing a nonvolatile solute is higher than the boiling point of a pure solvent.

Short Answer

Expert verified
A solution containing a nonvolatile solute has a higher boiling point because the solute lowers the solvent's vapor pressure, requiring a higher temperature to match atmospheric pressure.

Step by step solution

01

Understanding Boiling Point Elevation

Boiling point elevation is the phenomenon where the boiling point of a solution is higher than that of the pure solvent. This occurs when a nonvolatile solute is added to a solvent, causing the boiling point to increase.
02

Recognizing the Role of Vapor Pressure

In a pure solvent, the boiling point is reached when its vapor pressure equals the atmospheric pressure. However, the presence of a nonvolatile solute lowers the vapor pressure of the solvent, meaning the solution must be heated to a higher temperature for its vapor pressure to reach atmospheric pressure.
03

Colligative Properties and the Solute's Effect

Boiling point elevation is a colligative property, meaning it depends on the number of solute particles rather than their identity. The solute particles interfere with the solvent molecules' ability to escape into the vapor phase, necessitating a higher temperature to achieve boiling.
04

Formula for Boiling Point Elevation

The change in boiling point can be calculated using the formula: \[\Delta T_b = i imes K_b imes m\]where \( \Delta T_b \) is the boiling point elevation, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant of the solvent, and \( m \) is the molality of the solution.

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Key Concepts

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

Colligative Properties
When discussing boiling point elevation, colligative properties come into play. These properties depend on the number of solute particles in a solution, rather than the kind of particles. This means that whether you add sugar or salt to water, the effect on the boiling point primarily depends on how many particles dissolve鈥攁nd not their chemical nature.
  • Colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.
  • They are vital in understanding how solutions behave compared to their pure solvents.
Adding a solute like salt or sugar to water results in fewer water molecules at the surface. This interference requires more energy鈥攁nd thus a higher temperature鈥攆or the liquid to transition into the vapor phase.
Vapor Pressure
The concept of vapor pressure is essential to understanding why boiling point elevation occurs. Vapor pressure is the pressure a liquid's vapor exerts when the liquid and vapor are in equilibrium. In a pure solvent, the boiling point is the temperature at which its vapor pressure equals atmospheric pressure.
  • Adding a nonvolatile solute lowers the solvent's vapor pressure.
  • Lower vapor pressure means more heat is needed for bubbles to form, leading to a higher boiling point.
When a nonvolatile solute is added, fewer solvent molecules can escape into the vapor phase. Thus, achieving atmospheric pressure requires a higher temperature, illustrating why solutions boil at a higher temperature than pure solvents.
Nonvolatile Solute
A nonvolatile solute plays a central role in the boiling point elevation effect. By its very nature, a nonvolatile solute does not easily evaporate, meaning it significantly alters the physical properties of a solvent when dissolved.
For boiling point elevation:
  • The presence of the solute decreases the number of solvent molecules at the liquid's surface.
  • Since fewer solvent molecules can escape, the vapor pressure stays lower until the temperature increases.
Thus, these solutes lead to an elevation of the boiling point because more heat is required to reach the necessary vapor pressure to boil.
Ebullioscopic Constant
The ebullioscopic constant, denoted as \( K_b \), is a crucial factor in calculating boiling point elevation. This constant gives us the change in boiling point per one molal increase in solute concentration.
The formula \( \Delta T_b = i \times K_b \times m \) illustrates how to quantify the increase in boiling point temperatures:
  • \( \Delta T_b \) is the change in boiling point.
  • \( i \) is the van't Hoff factor, representing the degree of solute dissociation.
  • \( m \) is the molality of the solution.
Understanding the ebullioscopic constant helps in predicting how much the boiling point will elevate when a solute is added, making it a valuable tool in both academic and practical chemistry settings.

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Most popular questions from this chapter

A \(55-\mathrm{g}\) sample of a gaseous fuel mixture contains \(0.51\) mole fraction propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\); the remainder of the mixture is butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\). What are the masses of propane and butane in the sample?

Even though the oxygen demands of trout and bass are different, they can exist in the same body of water. However, if the temperature of the water in the summer gets above about \(23^{\circ} \mathrm{C}\), the trout begin to die, but not the bass. Why is this the case?

In a mountainous location, the boiling point of pure water is found to be \(95^{\circ} \mathrm{C}\). How many grams of sodium chloride must be added to \(1 \mathrm{~kg}\) of water to bring the boiling point back to \(100^{\circ} \mathrm{C}\) ? Assume that \(i=2\).

A natural gas mixture consists of \(90.0\) mole percent \(\mathrm{CH}_{4}\) (methane) and \(10.0\) mole percent \(\mathrm{C}_{2} \mathrm{H}_{6}\) (ethane). Suppose water is saturated with the gas mixture at \(20^{\circ} \mathrm{C}\) and \(1.00\) atm total pres sure, and the gas is then expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) at \(20^{\circ} \mathrm{C}\) and \(1.00\) atm are \(0.023 \mathrm{~g} / \mathrm{L} \mathrm{H}_{2} \mathrm{O}\) and \(0.059 \mathrm{~g} / \mathrm{L} \mathrm{H}_{2} \mathrm{O}\), respectively.

Ten grams of the hypothetical ionic compounds \(\mathrm{XZ}\) and YZ are each placed in a separate \(2.0-\mathrm{L}\) beaker of water. \(\mathrm{XZ}\) completely dissolves, whereas YZ is insoluble. The energy of hydration of the \(\mathrm{Y}^{+}\) ion is greater than the \(\mathrm{X}^{+}\) ion. Explain this difference in solubility.
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