CHM 2046C Module 8 Part II Sample Exam Answers

Part S: Boiling Point Elevation Problem 4 points

A solution of glycerol, C₃H₅(OH)₃, in 735 g of water has a boiling point of 104.4 ^oC at a pressure of 760 mm Hg. What is the mass of glycerol in the solution? Assume the solute to be a nonvolatile solute and the k_bp of water is + 0.52 ^oC/m .

104.4 ^oC

-100.0 ^oC

-------------

4.4 ^oC = ΔT

ΔT = k_bp m

4.4 ^oC = (+0.52 ^oC/m ) m

m = 8.46 m

Answer:

735 g H2O		8.46 mole Gly		92.0 g GLY
-------------	X	-------------	X	------------------	=	572.2 g GLY
1		1000.0 g H2O		1 mol Gly

What is the mole fraction of the SOLUTE?

572.2 g GLY solute		1 mol Gly solute
---------------	X	--------------	=	6.21 mol Gly solute
1		92.0 g Gly solute

735 g H2O		1 mol H2O
---------------	X	--------------	=	40.83 mol H2O Solvent
1		18 g H2O

Total Mole_solution = 6.21 mol _{GLY solute} + 40.83 mol _{H2O solvent} = 47.04 mol

Answer:

		6.21 mol H2O Solvent
Χsolvent	=	--------------	=	0.132
		47.04 mol solution

See Example 14.6 p676

Work example 14.6 p677

More sample Boiling Point Elevation problems may be found at:

Kotz 6^th Chap 14 p 693 #29-34

Part T: Freezing Point Depression Problem 4 points

What mass of ethylene glycol, HOCH₂CH₂OH, must be added to 6.50 Kg of water to lower the freezing point of the water from 0.0 to -10.0 ^oC. (This is supersize car radiator) Assume the solute to be a nonvolatile solute and the k_fp of water is - 1.86 ^oC/m .

-10.0 ^oC

- 0.0 ^oC

-------------

-10.0 ^oC = ΔT

ΔT = k_fp m

-10.00 ^oC = (-1.86 ^oC/m ) m

m = 5.38 m

Answer:

6.50 kg H2O		5.38 mole Glycol		62.0 g Glycol
-------------	X	-------------	X	------------------	=	2168 g Glycol
1		1.0 kg H2O		1 mol Glycol

See Example 14.7 p 678

Work Exercise 14.7 p 678

More sample Freezing Point Depression problems may be found at: Kotz 6^th Chap 14 p 694 #35-38