SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 7

Liquids, Solids, and Phase Changes

DILUTION

In dilution problems, the expression of molarity gives the quantity of solute per volume of solution. The amount of solute dissolved in a given volume of solution is equal to the product of the concentration times the volume. Hence, 0.5 liter of 2 M solution contains

M ×V = amount of solute (in moles)

Notice that volume units must be identical.

If you dilute a solution with water, the amount or number of moles of solute present remains the same, but the concentration changes. You can use the expression:

Before   After
M1V1 = M2V2

This expression is useful in solving problems involving dilution.

TIP 

Use this equation for dilution problems:
M1V1 = M2V2
1 = solution before
2 = solution after

EXAMPLE: If you wish to make 1 liter of solution that is 6 M into 3 M solution, how much water must be added?

M1V1 = M2V2
6 M × 1 L = 3 M × x L

Solving this expression:
x L = 2 L. This is the total volume of the solution after dilution and means that 1 liter of water had to be added to the original volume of 1 liter to get a total of 2 liters for the dilute solution volume.

An important use of the molarity concept is in the solution of titration problems, which are covered in Chapter 11, along with pH expressions of concentration for acids.