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Chapter 12: Problem 102

Which aqueous solution has the lower boiling point, \(0.10 \mathrm{~m} \mathrm{NaBr}\) or \(0.10 \mathrm{~m} \mathrm{CaCl}_{2}\) ?

Short Answer

Expert verified
The solution with the lower boiling point is 0.10 m NaBr.

Step by step solution

01

Understand the Concept of Boiling Point Elevation

The boiling point of a solution is elevated when a solute is dissolved in a solvent. The extent of boiling point elevation depends on the number of solute particles in solution (the van 't Hoff factor).
02

Determine the Van 't Hoff Factor

The van 't Hoff factor (f) is the number of particles into which a compound dissociates in solution. \(\text{NaBr}\) dissociates into 2 ions: Na\(^+\) and Br\(^-\), so \(i = 2\). \(\text{CaCl}_2\) dissociates into 3 ions: Ca\(^{2+}\) and 2 Cl\(^-\), so \(i = 3\).
03

Calculate the Boiling Point Elevation

The boiling point elevation can be calculated using the formula: \(\Delta T_b = iK_bm\), where \(K_b\) is the ebullioscopic constant and \(m\) is the molality of the solution. Since \(m\) and \(K_b\) are the same for both solutions, the difference in \(\Delta T_b\) is due to the difference in \(i\).
04

Compare the Boiling Points

The higher the \(i\), the greater the elevation in boiling point. Thus, \(0.10 \mathrm{~m} \mathrm{CaCl}_2\) with \(i = 3\) will have a greater boiling point elevation compared to \(0.10 \mathrm{~m} \mathrm{NaBr}\) with \(i = 2\). Therefore, \(0.10 \mathrm{~m} \mathrm{NaBr}\) has the lower boiling 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 is an essential concept when discussing boiling point elevation. It represents the number of individual particles into which a compound dissociates when it dissolves in a solvent. In simpler terms, it's a way to count the number of ions or molecules that result from dissolving a solute. For example,
  • NaBr dissociates into Na\(^+\) and Br\(^-\), thus it has a van 't Hoff factor \(i = 2\).
  • Similarly, CaCl\(_2\) dissociates into one Ca\(^{2+}\) and two Cl\(^-\), giving it \(i = 3\).
These values indicate how many particles are produced in solution, and consequently determine the extent to which these particles affect boiling point elevation.
Ebullioscopic constant
The ebullioscopic constant, represented as \(K_b\), is a property specific to each solvent, and it quantifies how much the boiling point of the solvent will increase per molal concentration of solute particles. It is a critical factor when calculating the change in boiling point using the formula:\[\Delta T_b = iK_bm\]Here, \(\Delta T_b\) is the change in boiling point, \(i\) is the van 鈥檛 Hoff factor, \(m\) is the molality, and \(K_b\) is the ebullioscopic constant.

The ebullioscopic constant allows us to apply a proportional relationship between the number of solute particles and the elevation in boiling point, showcasing how each unique solvent will respond to added solute.
Solute dissociation
Solute dissociation is the process by which an ionic compound separates into its individual ions when dissolved in a solvent. This process is crucial in understanding colligative properties, such as boiling point elevation. Consider how:
  • NaBr, an ionic compound, dissociates into Na\(^+\) and Br\(^-\).
  • CaCl\(_2\), another ionic compound, dissociates into one Ca\(^{2+}\) and two Cl\(^-\).
The degree to which a solute dissociates impacts the van 't Hoff factor, influencing how much the solute affects the boiling point of the solution.

Understanding solute dissociation helps predict the changes in boiling point and is fundamental when comparing different solutions for properties like boiling and freezing point alterations.
Molality
Molality is a measure of the concentration of a solution, calculated as the moles of solute per kilogram of solvent. Unlike molarity, molality does not depend on temperature or pressure changes, making it ideal for calculations involving boiling point and freezing point phenomena. It is symbolized as \(m\).

When analyzing boiling point elevation, molality plays a key role:
  • For both the NaBr and CaCl\(_2\) solutions, molality \(m = 0.10\, \text{mol/kg}\) remains the same.
  • This allows us to focus on how the van 't Hoff factor, not molality, impacts the boiling point.
Molality provides a stable foundation for carrying out calculations involving colligative properties, leading to more accurate and reliable results.

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Most popular questions from this chapter

What mass of solution containing \(6.50 \%\) sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) by mass contains \(1.75 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4} ?\)

Arrange the following substances in order of increasing solubility in hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}: \mathrm{CH}_{2} \mathrm{OHCH}_{2} \mathrm{OH},\) \(\mathrm{C}_{10} \mathrm{H}_{22}, \mathrm{H}_{2} \mathrm{O}\)

A green leafy salad wilts if left too long in a salad dressing containing vinegar and salt. Explain what happens.

Equal numbers of moles of two soluble, substances, substance A and substance \(\mathrm{B}\), are placed into separate 1.0 - \(\mathrm{L}\) samples of water. a. The water samples are cooled. Sample A freezes at \(-0.50^{\circ} \mathrm{C},\) and Sample \(\mathrm{B}\) freezes at \(-1.00^{\circ} \mathrm{C}\). Explain how the solutions can have different freezing points. b. You pour \(500 \mathrm{~mL}\) of the solution containing substance B into a different beaker. How would the freezing point of this 500 -mL portion of solution \(\mathrm{B}\) compare to the freezing point of the 1.0 - \(\mathrm{L}\) sample of solution \(\mathrm{A}\) ? c. Calculate the molality of the solutions of \(\mathrm{A}\) and \(\mathrm{B}\). Assume that \(i=1\) for substance \(\mathrm{A}\). d. If you were to add an additional \(1.0 \mathrm{~kg}\) of water to solution \(\mathrm{B}\), what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula.

What is the vapor pressure at \(23^{\circ} \mathrm{C}\) of a solution of \(1.20 \mathrm{~g}\) of naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\), in \(25.6 \mathrm{~g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) ? The vapor pressure of pure benzene at \(23^{\circ} \mathrm{C}\) is \(86.0 \mathrm{mmHg} ;\) the vapor pressure of naphthalene can be neglected. Calculate the vapor-pressure lowering of the solution.
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