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The Colligative Properties

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Sections
Solutions, Molarity and Intermolecular Forces	7 mins	0 completed	Learn
Henry's Law	20 mins	0 completed	Learn Summary
Calculate Molarity	27 mins	0 completed	Learn Summary
Mass Percent	10 mins	0 completed	Learn
Molality	15 mins	0 completed	Learn
Mole Fraction	14 mins	0 completed	Learn
The Colligative Properties	45 mins	0 completed	Learn

Additional Practice
Making Solutions
Freezing Point Depression
Colloid

Additional Guides
ppm
The Freezing Point Depression
Boiling Point Elevation

Learn Additional Practice

Solution: Rank the following aqueous solutions in order of decreasing boiling point.A. 0.020 m NaBr > 0.012 m KCl > 0.005 m SrCl2 > 0.012 m sucroseB. 0.020 m NaBr > 0.005 m SrCl2 > 0.012 m KCl =

Problem

Rank the following aqueous solutions in order of decreasing boiling point.

A. 0.020 m NaBr > 0.012 m KCl > 0.005 m SrCl2 > 0.012 m sucrose

B. 0.020 m NaBr > 0.005 m SrCl2 > 0.012 m KCl = 0.012 m sucrose

C. 0.005 m SrCl2 > 0.012 m KCl > = 0.012 m sucrose > 0.020 m NaBr

D. 0.012 m sucrose > 0.012 m KCl > 0.005 m SrCl2 > 0.020 m NaBr

The Colligative Properties

Solution: Rank the following aqueous solutions in order of decreasing boiling point.A. 0.020 m NaBr &gt; 0.012 m KCl &gt; 0.005 m SrCl2 &gt; 0.012 m sucroseB. 0.020 m NaBr &gt; 0.005 m SrCl2 &gt; 0.012 m KCl =

Problem

Practice Problems

Solution: Rank the following aqueous solutions in order of decreasing boiling point.A. 0.020 m NaBr > 0.012 m KCl > 0.005 m SrCl2 > 0.012 m sucroseB. 0.020 m NaBr > 0.005 m SrCl2 > 0.012 m KCl =