Ch.12 - SolutionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: The solubility of Cr ( NO3 )3 9 H2O in water is 208 g per 100 g of water at 15 oC. A solution of Cr ( NO3 )3 9 H2O in water at 35 oC is formed by dissolving 313 g in 100 g water. When this solution

Problem

The solubility of Cr ( NO3 )3 9 H2O in water is 208 g per 100 g of water at 15 oC. A solution of Cr ( NO3 )3 9 H2O in water at 35 oC is formed by dissolving 313 g in 100 g water. When this solution is slowly cooled to 15 oC, no precipitate forms.

At equilibrium, what mass of crystals do you expect to form?