Page 179 - Theory and Problems of BEGINNING CHEMISTRY

168 MOLARITY [CHAP. 11

11.5. How can you make 2.25 L of 4.00 M sugar solution?
Ans. The solution will contain 9.00 mol sugar:
4.00 mol

2.25 L = 9.00 mol
1L
Thus, place 9.00 mol sugar in a liter or two of water, mix until dissolved, dilute the resulting solution to
2.25 L, and mix thoroughly.

11.6. Calculate the volume of 1.75 M solution required to contain 4.20 mol of solute.
1L

Ans. 4.20 mol = 2.40 L
1.75 mol
11.7. A 10.0-mL solution contains 2.40 mmol of solute. What is its molarity?

Ans. 2.40 mmol = 0.240 M
10.0mL
Molarity can be calculated by dividing millimoles by milliliters.

11.8. How many moles of solute are present in 29.4 mL of 0.606 M solution?
Ans. (0.0294 L)(0.606 mol/L) = 0.0178 mol

11.9. Calculate the number of milliliters of 1.25 M solution required to contain 0.622 mol of solute.
1L

Ans. 0.622 mol = 0.498 L = 498 mL
1.25 mol
11.10. What volume of 1.45 M NaCl solution contains 71.3 g NaCl?
1 mol NaCl

Ans. 71.3 g NaCl = 1.22 mol NaCl
58.5 g NaCl
1 L solution

1.22 mol NaCl = 0.841 L = 841 mL
1.45 mol NaCl
11.11. What is the concentration of a solution prepared by dissolving 22.2 g NaCl in sufficient water to make
86.9 mL of solution?
Ans. Molarity is in moles per liter. We must change the grams of NaCl to moles and the milliliters of solution to
liters.
1 mol NaCl

22.2 g NaCl = 0.379 mol
58.5 g NaCl
0.379 mol/0.0869 L = 4.36 M

11.12. How many grams of NaCl are present in 72.1 mL of 1.03 M NaCl?
Ans. (0.0721 L)(1.03 mol/L) = 0.0743 mol
(0.0743 mol)(58.5 g NaCl/mol) = 4.35 g NaCl

11.13. Calculate the volume of 0.900 M solution required to contain 1.84 mol of solute.
1L

Ans. 1.84 mol = 2.04 L
0.900 mol
11.14. How many milligrams of NaOH are present in 35.0 mL of 2.18 M NaOH?
2.18 mmol 40.0mg

Ans. 35.0mL = 3050 mg
1mL 1 mmol