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CHEM103 / CHEM 103 Module 6:

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

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)23 Ca+2 + 2 PO4-31.86 x 0.100 x 5 = 0.93

2.) AlCl31 Al+3 + 3 Cl-1.86 x 0.100 x 4 = 0.744

3.) CaCl21 Ca+2 + 2 Cl-1.86 x 0.100 x 3 = 0.558

4.) KCl1 K+ + 1 Cl-1.86 x 0.100 x 2 = 0.372

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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 %

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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 figure

Answer:

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

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

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