Module 6 Exam Quiz
Module 6 Exam
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Categorize (1) milk, (2) whipped cream and (3) fog as emulsion, foam or aerosol.
Correct Answer
- Milk is an emulsion (liquid in liquid),
- Whipped cream is a foam (gas in liquid)
- Fog is an aerosol (liquid in gas).
Explain how and why the presence of a solute affects the freezing point of a solvent.
Correct Answer
The presence of a solute lowers the freezing point of a solvent by forcing solvent molecules away from the growing solid crystal. In order for the solvent molecules to reach the crystal and add themselves to the freezing solid, they must be slowed down to a lower kinetic energy by lowering the temperature.
Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freezing point to highest freezing point.
MgI2 ¸AlCl3, Mg3(PO4)2, NaI
Correct Answer
MgI2 → Mg+2 + 2 I- ∆tf = 1.86 x 0.1 x 3 = 2nd lowest FP
AlCl3 → Al+3 + 3 Cl- ∆tf = 1.86 x 0.1 x 4 = 3rd lowest FP
Mg3(PO4)2 → 3 Mg+2 + 2 PO4-3 ∆tf = 1.86 x 0.1 x 5 = lowest FP
NaI → Na+ + I- ∆tf = 1.86 x 0.1 x 2 = highest FPFP: Mg3(PO4)2 < AlCl3 < MgI2 < NaI
Show the calculation of the mass percent solute in a solution of 20.8 grams of Ba(NO3)2 in 400 grams of water. Report your answer to 3 significant figures.
Correct Answer
Mass % = (gsolute / gsolute + gsolvent) x 100%
Mass % = (20.8 / 20.8 + 400) x 100 = 4.94%
Show the calculation of the molality of a solution made by dissolving 24.5 grams of C4H8O4 in 400 grams of water. Report your answer to 3 significant figures.
Correct Answer
Molar mass of C4H8O4=4(12.01)+8(1.008)+4(16.00)=120.104 g/mol
# moles of solute=24.5 g/120.10 g/mol=0.204 mol
Molality= # moles/ kg of solvent = 0.204 mol/(400 g*1kg/1000 g)= 0.510 m
Show the calculation of the molarity of a solution made by dissolving 35.9 grams of Mg(NO3)2 to make 400 ml of solution. Report your answer to 3 significant figures.
Correct Answer
Molar mass of Mg(NO3)2=24.31+2(14.01)+6(16.00)= 148.33 g/mol
Molarity = (gsolute / MW) / (mlsolvent / 1000)
Molarity = (35.9 / 148.33) / (400 / 1000) = 0.605 M
Show the calculation of the mass of Ca3(PO4)2 needed to make 350 ml of a 0.150 M solution. Report your answer to 3 significant figures.
Correct Answer
Molarity = (moles) / (mlsolvent / 1000)
0.150 = (moles) / (350 / 1000)
Moles = 0.150 x 0.350 = 0.0525
Moles = (gsolute / MW)
0.0525 = (gsolute / 310.18)
gsolute = 0.0525 x 310.18 = 16.3 g
Show the calculation of the volume of 0.827 M solution which can be prepared using 26.4 grams of Ca3(PO4)2.
Correct Answer
molessolute = gsolute / MW
molessolute = 26.4 g / 310.18 = 0.0851 mol
Molarity = moles / (mL /1000)
0.827 = 0.0851 / (mL / 1000)
mL / 1000 = 0.0851 / 0.827 = 0.1029
mL = 0.1029 x 1000 = 103 mL
Show the calculation of the freezing point of a solution made by dissolving 15.6 grams of the nonelectrolyte C3H6O3 in 200 grams of water. Kf for water is 1.86, freezing point of pure water is 0oC. Calculate your answer to 0.01oC.
Correct Answer
molality = (gsolute / MW) / (gsolvent / 1000)
molality = (15.6 / 90.078) / (200 / 1000) = 0.8659 m
∆tf = Kf x m = 1.86 x 0.8659 = 1.61oC
FPsolution = FPsolvent - ∆tf = 0oC - 1.61 = - 1.61oC
Show the calculation of the molar mass (molecular weight) of a solute if a solution of 14.6 grams of the solute in 200 grams of water has a freezing point of -1.35oC. Kf for water is 1.86 and the freezing point of pure water is 0oC. Calculate your answer to 0.1 g/mole.
Correct Answer
∆tf = Kf x m
molality = ∆tf / Kf = 1.35 / 1.86 = 0.726 m
molality = (gsolute / MW) / (gsolvent / 1000)
0.726 = (moles) / (200 / 1000)
Moles = 0.726 x 0.200 = 0.1452
0.1452 = (14.6 / MW)
MW = 14.6 / 0.1452 = 100.6 g/mol
