CHEM103 CHEM 103 Final Exam:

CHEM103 / CHEM 103 Final Exam:

(Latest Update 2025 / 2026) General Chemistry I with Lab | Questions & Answers | Grade A | 100% Correct - Portage Learning

Question:

Arrange the following compounds in a vertical list from highest boiling point (top) to lowest boiling point (bottom) and explain your answer on the basis of whether the substance is Polar, Nonpolar, Ionic, Metallic or Hydrogen

bonding: Mg, H2O, Ne, HCl, LiCl

Answer:

1.) Mg (metallic) & LiCl (ionic) 2.) H2O (hydrogen bonding) 3.) HCl (polar) 4.) Ne (nonpolar)

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Question:

Air is a homogeneous mixture solution made up of 20% oxygen gas and 80% nitrogen gas. Which gas is the solvent and which is the solute. Explain your answer.

Answer:

Oxygen is the solute and nitrogen is the solvent because nitrogen is the most abundant substance in the air.

Question:

Explain how and why the presence of a solute affects the freezing point of a solvent.

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.

Question:

Show the calculation of the mass percent solute in a solution of 22.6 grams of Al2(SO4)3 in 400 grams of water. Report your answer to 3 significant figures.

Answer:

Mass % = g of solute / (g of solute + g of solvent) x 100 Mass % = 22.6 g / (22.6 g + 400 g) x 100 Mass % = 5.35 % 2 / 4

Question:

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.AlCl3, Ca3(PO4)2, KCl, CaCl2

Answer:

1.) Ca3(PO4)2

• Ca+2 + 2 PO4-3

1.86 x 0.100 x 5 = 0.93

2.) AlCl3

• Al+3 + 3 Cl-

1.86 x 0.100 x 4 = 0.744

3.) CaCl2

• Ca+2 + 2 Cl-

1.86 x 0.100 x 3 = 0.558

4.) KCl

• K+ + 1 Cl-

1.86 x 0.100 x 2 = 0.372

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Question:

Show the calculation of the molality of a solution made by dissolving 25.5 grams of C5H10O5 in 350 grams of water. Report your answer to 3 significant figures.

Answer:

m = (g of solute / MW of solute) / (g of solvent / 1000) m = (25.5 g / 150.13 g/mol) / (350 g / 1000) m = 0.485 m

Question:

Show the calculation of the molarity of a solution made by dissolving 32.4 grams of Ba(NO3)2 to make 300 ml of solution. Report your answer to 3 significant figures.

Answer:

Molarity = (gsolute / MW) / (mlsolvent / 1000)

Molarity = (32.4 / 261.55) / (300 / 1000) = 0.413 M

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