CHM 2046C Sample Exam Answers

Module Eight Part II: Solutions Chapter 14

Part O: Concentration Calculations 12 points

You want to prepare an aqueous solution of glycerol, C₃H₅(OH)₃ in which the mole fraction of the solute is 0.093. What mass of glycerol must you add to 425 g of water to make the solution? What is the molality of the solution? What is the weight percent?

Mole Fraction of Solvent:

1.000 moles _solution – 0.093 moles _{GLY solute} = 0.907 mole _{H2O solvent}

Mass of Glycerol

425 g H2O		1 mole H2O		0.093 mol GLY		92.0 g GLY
-------------	X	-------------	X	------------------	X	--------------	=	222.7 g GLY
1		18.0 g H2O		0.907 mol H2O		1 mol GLY

Molality

222.7 g _GLY/ 92.0 g/mol = 2.42 mol _GLY

2.42 mol GLY
---------------	=	5.70 molal
0.425 kg H2O

Mass Percent

222.7 g _GLY + 425 g _H2O = 647 g _solution

222.7 g GLY solute		100
---------------	X	-----	=	33.4 g GLY solute /100 g solution or 33.4 % GLY
647 g solution		100

McMurray Work mass % Examples 11.2 and 11.3 p 406

Try McMurray Problems mass % 11.3, 11.4, 11.5 p 406

See McMurray Molality Worked Examples 11.4 p407 11.5, 11.6 p 408

Work Molality Problems 11.6, 11.7, 11.8, 11.9, 11.10 p 409

End of Chapter Exercises: Units of Concentration (Section 11.3)

Lots of exercises: 11.48-11.69 p433-434

See Kotz Example 14.1 p 661; Example 14.2 p661

Work Example 14.1 and 14.2 p662

Similar Problems: Kotz 6^th Chapter 14 p692 Q 1-12

Additional Solved Problems from Another Book: