Skip to main content
Pearson+ LogoPearson+ Logo
Ch.13 - Properties of Solutions
All textbooksBrown 14th EditionCh.13 - Properties of SolutionsProblem 71
Chapter 13, Problem 71

List the following aqueous solutions in order of increasing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m Zn(NO3)2.

Verified step by step guidance
1
Identify the colligative property involved: Boiling point elevation is a colligative property, which depends on the number of solute particles in a solution.
Determine the van 't Hoff factor (i) for each solute: Glucose (C_6H_{12}O_6) is a non-electrolyte, so i = 1. LiBr is an electrolyte that dissociates into Li^+ and Br^-, so i = 2. Zn(NO_3)_2 dissociates into Zn^{2+} and 2 NO_3^-, so i = 3.
Calculate the effective molality for each solution: Multiply the molality (m) by the van 't Hoff factor (i) for each solution. For glucose, it's 0.120 m * 1. For LiBr, it's 0.050 m * 2. For Zn(NO_3)_2, it's 0.050 m * 3.
Compare the effective molalities: The solution with the highest effective molality will have the highest boiling point, and the one with the lowest effective molality will have the lowest boiling point.
List the solutions in order of increasing boiling point based on their effective molalities.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the identity of the solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. In this context, the boiling point elevation is crucial for determining the order of the solutions based on their molal concentrations.
Recommended video:
Guided course
01:26
Colligative Properties

Boiling Point Elevation

Boiling point elevation occurs when a non-volatile solute is added to a solvent, resulting in an increase in the boiling point of the solution compared to the pure solvent. The extent of this elevation is directly proportional to the molality of the solute and the van 't Hoff factor (i), which accounts for the number of particles the solute dissociates into. Thus, solutions with higher concentrations of solute particles will have higher boiling points.
Recommended video:
Guided course
03:05
Boiling Point Elevation

Van 't Hoff Factor (i)

The van 't Hoff factor (i) represents the number of particles into which a solute dissociates in solution. For example, glucose does not dissociate (i=1), while LiBr dissociates into two ions (Li+ and Br-, i=2), and Zn(NO3)2 dissociates into three ions (Zn2+ and two NO3-, i=3). This factor is essential for calculating the effective concentration of solute particles, which influences the boiling point elevation of the solution.
Recommended video:
Guided course
02:50
Van't Hoff Factor
Related Practice
Open Question

At 20 °C, the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of 35 torr at 20 °C?

Textbook Question

At 20 °C, the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

2271
views
1
comments
Textbook Question
(a) Does a 0.10 m aqueous solution of NaCl have a higher boiling point, a lower boiling point, or the same boiling point as a 0.10 m aqueous solution of C6H12O6?
1109
views
Open Question
List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m KBr, 0.030 m phenol (C6H5OH).
Textbook Question

Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol (C3H8O3) in ethanol, (b) 0.240 mol of naphthalene (C10H8) in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g of glucose (C6H12O6) in 188 g of water.

1130
views
Textbook Question

Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.25 m glucose in ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and 0.15 mol KBr in 150 g H2O.

1576
views