Lesson 11.2 Concentration
Suggested Reading
 Zumdahl Chapter 11 Section11.1
Essential Questions
 In what ways can concentration be expressed?
Learning Objectives


Define molality.
Solve problems relating to the mass percent, mole fraction, and molality of a solution.
Introduction
The concentration of a solute is the amount of solute dissolved in a given quantity of solvent or solution. The quantity of solvent or solution can be expressed in terms of volume or in terms of molar amount. Thus, there are different ways of expressing the concentration of a solution.
Molarity
Recall, the molarity ( M ) of a solution is the moles of solute in liter of solution.
M = moles of solute/liters of solution
For example, 0.20 mol of ethylene glycol dissolved in enough water to give
2.0 L of solution has a molarity of

0.20 mol ethylene glycol / 2.0 L of solution = 0.10 M ethylene glycol
This unit is especially useful when you wish to dispense a given amount of solute, because the unit directly relates the amount of solute to the volume of solution.
Other concentration units are defined in terms of the mass or molar amount of solvent or solution. The most important of these are mass percent of solute, molality, and mole fraction.
Mass Percent of Solute
The mass percent of a solute is the percentage by mass of solute contained in a solution.
For example, an aqueous solution that is 3.5% sodium chloride by mass contains 3.5 g of NaCl in 100.0 g of solution. It could be prepared by dissolving 3.5 g of NaCl in 96.5 g of water.
Example: Mass Percent of Solute
How would you prepare 425 g of an aqueous solution containing 2.40% sodium acetate by mass, NaC
2
H
3
O
2
?
Solution:
The mass of sodium acetate in 425 g of solution is
425 g x 0.0240 = 10.2 g
The quantity of water in the solution is 425 g - 10.2 g = 415 g. You would prepare the solution by dissolving 10.2 g of sodium acetate in 415 g of water.
Molality

The molality (m) of a solution is the moles of solute per kilogram of solvent.
For example, 0.20 mol of ethylene glycol dissolved in 2.0 kg of solvent (2.0 x 10 3 g) of water has a molality of
The units of molality and molarity are easy to confuse. Note that molality is defined in terms of the mass of solvent, and molarity is defined in terms of the volume of solution. Molality is used when the temperature of the solution is variable. Since molarity depends on the volume of the solution, it changes slightly with temperature. Molality does not, since it depends on mass.
Example: Calculating the Molality of Solute
Glucose, C
6
H
12
O
6
, is a sugar that occurs in fruits. It is also known as "blood sugar" because it is found in blood and is the body's main source of energy. What is the molality of a solution containing 5.67 g of glucose dissolved in 25.2 g of water?
Solution:
The moles of glucose (M = 180.2 g/mol) in 5.67 g are
The mass of water is 25.2 g, or 25.2 x 10 -3 kg, so the molality is

Mole Fraction
The mole fraction of a component substance A (X
A
) in a solution is defined as the moles of component substance divided by the total moles of solution, where the total moles of solution is the moles solute plus the moles of solvent.
For example, if a solution is made up of 1 mol of ethylene glycol and 9 mol of water, the total moles of solution are 10 mol. The mole fraction of ethylene glycol is 1/10 = 0.1, and the mole fraction of water is 9/10, or 0.9.
Multiplying mole fractions by 100 gives mole percent. Hence, the solution is
10 mole percent ethylene glycol and 90 mole percent of water. The sum of all mole fractions must equal 1. Use this fact to check your work.
Example: Calculating the Mole Fractions of Components
What are the mole fractions of glucose and water in a solution containing
5.67 g of glucose dissolved in 25.2 g of water.
Solution:
This is the glucose solution as in the previous example. From this problem we know that 5.67 g of glucose equals 0.0315 mole of glucose. The moles of water are
The total moles of solution are 1.40 mol + 0.0315 mol = 1.432 mol.
The mole fraction of glucose:

The mole fraction of water:
HOMEWORK: Finish book questions assigned for lesson 11.1.
Practice exercises 13.2, 13.3, 13.5, 13.6, 13.7