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List the following aqueous solutions in order of increasing boiling point: 0.120 \(\mathrm{m}\) glucose, 0.050 \(\mathrm{m}\) LiBr, 0.050 \(\mathrm{m}\) \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\)

Short Answer

Expert verified
The aqueous solutions are ranked in order of increasing boiling point as follows: 0.120 m glucose < 0.050 m LiBr < 0.050 m \(Zn(NO_3)_2\).

Step by step solution

01

Understanding Boiling Point Elevation

Boiling point elevation is a colligative property, meaning it depends only on the number of solute particles in the solution, not their identity. The formula for boiling point elevation is: \[\Delta T_b = K_b \cdot m \cdot i\] Where \(\Delta T_b\) is the change in boiling point, \(K_b\) is the molal boiling point elevation constant for the solvent (water in this case), \(m\) is the molality of the solute, and \(i\) is the van't Hoff factor (number of particles formed by one formula unit of the solute).
02

Calculate the Boiling Point Elevation for Each Solution

We will calculate the boiling point elevation for each solution using the given molality and respective van't Hoff factors for each solute. For glucose (C₆H₁₂O₆), the formula unit doesn't dissociate in water, so its van't Hoff factor (i) is 1. For LiBr, the formula unit dissociates into two ions (Li⁺ and Br⁻), so its van't Hoff factor (i) is 2. For Zn(NO₃)₂, the formula unit dissociates into three ions (Zn²⁺, 2NO₃⁻), so its van't Hoff factor (i) is 3. Now, we can calculate the boiling point elevation for each solution: \[\Delta T_{b, glucose} = K_b \cdot m_{glucose} \cdot i_{glucose}\] \[\Delta T_{b, LiBr} = K_b \cdot m_{LiBr} \cdot i_{LiBr}\] \[\Delta T_{b, Zn(NO_3)_2} = K_b \cdot m_{Zn(NO_3)_2} \cdot i_{Zn(NO_3)_2}\] Using the given molality and calculated van't Hoff factors: \[\Delta T_{b, glucose} = K_b \cdot 0.120 \cdot 1\] \[\Delta T_{b, LiBr} = K_b \cdot 0.050 \cdot 2\] \[\Delta T_{b, Zn(NO_3)_2} = K_b \cdot 0.050 \cdot 3\]
03

Compare the Boiling Point Elevation and Rank the Solutions

We can compare the calculated boiling point elevations (note that the \(K_b\) for water will be the same in all cases and doesn't affect the ranking): \[\Delta T_{b, glucose} < \Delta T_{b, LiBr} < \Delta T_{b, Zn(NO_3)_2}\] Therefore, the aqueous solutions are ranked in order of increasing boiling point as follows: 0.120 m glucose < 0.050 m LiBr < 0.050 m Zn(NO₃)₂

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Key Concepts

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

Colligative Properties
Colligative properties are unique characteristics of solutions that rely on the number of solute particles present, rather than their specific types. This fascinating aspect means that it doesn't matter if the solute is sugar or salt; what matters is the quantity of particles these substances offer in the solution. There are several colligative properties, but some of the most common include:
  • Boiling point elevation - The increase in the boiling point of the solvent when a solute is added.
  • Freezing point depression - The decrease in the freezing point of the solvent due to the addition of a solute.
  • Osmotic pressure - The pressure required to stop the flow of solvent into the solution through a semi-permeable membrane.
Understanding colligative properties is essential for solving problems like boiling point elevation, which is exactly what we are doing in this exercise. By focusing on the quantity of particles from dissolved substances, we can easily determine their effects on the boiling or freezing points of solutions.
Van't Hoff Factor
The van't Hoff factor, often symbolized as 'i', is used to describe how solute particles influence colligative properties. Essentially, it tells us how many particles a solute breaks into when dissolved. For example:
  • Non-electrolytes like glucose do not dissociate in solution, so their van't Hoff factor is 1.
  • Electrolytes dissociate into multiple ions. For instance, LiBr splits into Li⁺ and Br⁻, giving it a van't Hoff factor of 2.
  • A compound like Zn(NO₃)₂ breaks into Zn²⁺ and 2 NO₃⁻ ions, resulting in a van't Hoff factor of 3.
When calculating colligative properties, the van't Hoff factor is essential because it directly impacts the number of solute particles in the solution. More particles generally lead to a greater effect on properties like boiling point. The formula for boiling point elevation includes the van't Hoff factor because it represents the total impact of the solute particles in a solution.
Aqueous Solutions
An aqueous solution is one where water serves as the solvent. Water is an excellent solvent due to its ability to dissolve a wide range of substances, making it versatile and useful for many chemical processes. When discussing boiling point elevation, aqueous solutions are prevalent because water has a well-understood boiling point and reacts predictably as a solvent.
  • The presence of solutes in an aqueous solution can alter the boiling and freezing points of the water.
  • Water's characteristics as a solvent also make it ideal for studying colligative properties.
  • Due to its polar nature, water can break down various ionic and covalent compounds, which affects the boiling point elevation based on the solute's van't Hoff factor.
Being familiar with aqueous solutions allows us to explore problems like those in this exercise, where we determine the effect of different solutes on water's boiling point. This understanding is crucial to predict changes in solvent behavior when different substances are dissolved.

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Most popular questions from this chapter

Calculate the molarity of the following aqueous solutions: (a) 0.540 \(\mathrm{g}\) of Mg \(\left(\mathrm{NO}_{3}\right)_{2}\) in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{b}) 22.4 \mathrm{gof}\) \(\mathrm{LiClO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) in 125 \(\mathrm{mL}\) of solution, \((\mathrm{c}) 25.0 \mathrm{mL}\) of 3.50 \(\mathrm{M}\) \(\mathrm{HNO}_{3}\) diluted to 0.250 \(\mathrm{L}\)

Which of the following in each pair is likely to be more soluble in hexane, \(\mathrm{C}_{6} \mathrm{H}_{14} :(\mathbf{a}) \mathrm{CCl}_{4}\) or \(\mathrm{CaCl}_{2},(\mathbf{b})\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) or glycerol, \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH},(\mathbf{c})\) octanoic \(\mathrm{acid}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) or acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH} ?\) Explain your answer in each case.

The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in \(1.0 \times 10^{3}\) gal of seawater.

The density of acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is 0.786 \(\mathrm{g} / \mathrm{mL}\) and the density of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is 0.791 \(\mathrm{g} / \mathrm{mL} . \mathrm{A}\) solution is made by dissolving 22.5 \(\mathrm{mL}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) in 98.7 \(\mathrm{mL}\) of \(\mathrm{CH}_{3} \mathrm{CN}\) . (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in the solution?

Arrange the following aqueous solutions, each 10\(\%\) by mass in solute, in order of increasing boiling point: glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right),\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) sodium nitrate (\textrm{NaNO} _ { 3 } ) .

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