Question

In each pair, identify the solution that will have a higher boiling point. Explain. a. \(1.50\) moles of LiOH (strong electrolyte) or \(3.00\) moles of \(\mathrm{LiOH}\) (strong electrolyte) each in \(0.50 \mathrm{~L}\) of water. b. \(0.40 \mathrm{~mole}\) of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) (strong electrolyte) or \(0.40 \mathrm{~mole}\) of \(\mathrm{CsCl}\) (strong electrolyte) each in \(0.50 \mathrm{~L}\) of water.

Short answer

The solution with 3.00 moles of LiOH and the solution with Al(NO\textsubscript{3})\textsubscript{3} will have higher boiling points.

Step-by-step solution

– Understanding Boiling Point Elevation

Boiling point elevation is a colligative property, which means it depends on the number of solute particles in the solution and not on their identity. The formula for boiling point elevation is \[ \text{ΔT}_b = i \times k_b \times m \] where \( \text{ΔT}_b \) is the boiling point elevation, \( i \) is the van't Hoff factor (number of particles the compound dissociates into), \( k_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

– Problem (a) Calculation

For \( 1.50 \) moles of \( \text{LiOH} \) in \( 0.50 \mathrm{~L} \) of water:\( \text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1.50\text{~moles}}{0.50\text{~kg}} = 3.00\text{~mol/kg} \)The van’t Hoff factor for \( \text{LiOH} \) is \( 2 \) because it dissociates into \( \text{Li}^+ \) and \( \text{OH}^- \) ions.Thus, the effective number of particles (i × m) is \( 2 \times 3.00 = 6.00 \).For \( 3.00 \) moles of \( \text{LiOH} \) in \( 0.50 \mathrm{~L} \) of water:\( \text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{3.00\text{~moles}}{0.50\text{~kg}} = 6.00\text{~mol/kg} \)Since the van’t Hoff factor for \( \text{LiOH} \) is \( 2 \), the effective number of particles is \( 2 \times 6.00 = 12.00 \).Since the boiling point elevation increases with the number of particles, the solution with \( 3.00 \) moles of \( \text{LiOH} \) will have a higher boiling point.

– Problem (b) Calculation

For \( 0.40 \text{~mole} \) of \( \text{Al(NO}_3\text{)}_3 \) in \( 0.50 \mathrm{~L} \) of water:\( \text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.40 \text{~moles}}{0.50\text{~kg}} = 0.80\text{~mol/kg} \)The van’t Hoff factor for \( \text{Al(NO}_3\text{)}_3 \) is \( 4 \) as it dissociates into \( 1 \text{~Al}^{3+} \) ion and \( 3 \text{~NO}_3^- \) ions.Thus, the effective number of particles is \( 4 \times 0.80 = 3.20 \).For \( 0.40 \text{~mole} \) of \( \text{CsCl} \) in \( 0.50 \mathrm{~L} \) of water:\( \text{Molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.40\text{~moles}}{0.50\text{~kg}} = 0.80\text{~mol/kg} \)The van’t Hoff factor for \( \text{CsCl} \) is \( 2 \) because it dissociates into \( \text{Cs}^+ \) and \( \text{Cl}^- \) ions.Thus, the effective number of particles is \( 2 \times 0.80 = 1.60 \).Since the boiling point elevation increases with the number of particles, the solution with \( \text{Al(NO}_3\text{)}_3 \) will have a higher boiling point.

Key concepts

Colligative Properties

Colligative properties are special because they depend only on the number of solute particles in a solution and not on their identity. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. They are important in various scientific and industrial applications. For students, understanding colligative properties is crucial to mastering various chemistry concepts.

When it comes to boiling point elevation, it's all about how many solute particles are dissolved. More particles mean a higher boiling point. This happens because solute particles disrupt the solvent molecules, requiring more heat to turn the solution into vapor. Remember, it does not matter if the particles are sugar, salt, or anything else. It's the number that counts, not what they are made of.

This dependency makes colligative properties useful for determining molar masses and understanding the behavior of solutions. For example, knowing that more particles increase the boiling point helps in fields like cooking, where adding salt to water increases its boiling point, leading to faster cooking times.

van't Hoff Factor

The van't Hoff factor, denoted as \(i\), is a measure of how many particles a solute breaks into when it dissolves in a solvent. It's crucial for calculating colligative properties like boiling point elevation and freezing point depression.

For instance, if you dissolve sodium chloride (\text{NaCl}) in water, it dissociates into two particles: one \text{Na}^+ ion and one \text{Cl}^- ion, so \(i\) is 2. For a substance like \text{Al(NO}_3\text{)}_3, which dissociates into four particles (one \text{Al}^{3+} ion and three \text{NO}_3^- ions), \(i\) is 4.

The formula for boiling point elevation is \[ \text{ΔT}_b = i \times k_b \times m \] where:
- \(\text{ΔT}_b\) is the boiling point elevation,
- \(i\) is the van't Hoff factor,
- \(k_b\) is the ebullioscopic constant unique to each solvent,
- \(m\) is the molality of the solution.

To put it simply, the van't Hoff factor tells you how many particles a compound will produce in water. More particles mean greater changes in colligative properties. This is why \text{Al(NO}_3\text{)}_3, which has a higher \(i\), causes a more significant rise in boiling point than \text{CsCl}.

Molality

Molality (\(m\)) is a way to measure the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent. One reason molality is useful is that it does not change with temperature, unlike molarity which depends on the volume that can vary with temperature.

The formula for molality is: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

For example, if you have 1.50 moles of \text{LiOH} in 0.50 kg of water, the molality is \[ m = \frac{1.50 \text{ moles}}{0.50 \text{ kg}} = 3.00 \text{ mol/kg} \]

Molality becomes particularly important in calculating colligative properties like boiling point elevation using the formula \[ \text{ΔT}_b = i \times k_b \times m \] where all you need is the van’t Hoff factor, the ebullioscopic constant, and the molality.

Understanding molality is essential for solving problems related to boiling point elevation. Since molality measures how concentrated the particles are in a solution, it directly affects colligative properties. Higher molality means more particles per kilogram of solvent, leading to a higher boiling point.