Page 174 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 11] MOLARITY 163

1 L 2.0 mol

1 L 2.0 mol

1 L 2.0 mol

Fig. 11-1. 3.0 L of 2.0 M solution

EXAMPLE 11.3. What volume of 1.50 M solution contains 6.00 mol of solute?
1L

Ans. 6.00 mol = 4.00 L
1.50 mol
EXAMPLE 11.4. What is the molarity of a solution prepared by dissolving 63.0 g NaF in enough water to make 250.0 mL
of solution?
Ans. Molarity is defined in moles per liter. We convert each of the quantities given to those used in the definition:

1 mol NaF 1L
63.0 g NaF = 1.50 mol NaF 250.0mL = 0.2500 L
42.0 g NaF 1000 mL
1.50 mol
Molarity = = 6.00 M
0.2500 L
EXAMPLE 11.5. What is the concentration of a solution containing 1.23 mmol in 1.00 mL?

1 mol
Ans. 1.23 mmol
1000 mmol 1.23 mol
= = 1.23 M

1L 1L
1.00 mL
1000 mL
The numeric value is the same in mmol/mL as in mol/L. Thus, molarity can be defined as the number of millimoles
of solute per milliliter of solution, which is an advantage because chemists more often use volumes measured in
milliliters.
EXAMPLE 11.6. Do we know how much water was added to the solute in Example 11.1?
Ans. We have no way of knowing how much water was added. We know the final volume of the solution, but not the
volume of the solvent.
When water is added to a solution, the volume increases but the numbers of moles of the solutes do not
change. The molarity of every solute in the solution therefore decreases.

EXAMPLE 11.7. What is the final concentration of 2.00 L of 3.00 M solution if enough water is added to dilute the
solution to 10.0 L?
Ans. The concentration is the number of moles of solute per liter of solution, equal to the number of moles of solute
divided by the total volume in liters. The original number of moles of solute does not change:
3.00 mol

Number of moles = 2.00 L = 6.00 mol
1L