Ch.12 - SolutionsWorksheetSee all chapters
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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: How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?

Solution: How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?

Problem

How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?

Solution
  • Recall that molality can be calculated as:

  • Therefore, if we have 3.08 m of glycerin, this will appear as:

  • We need to calculate for the mass of C3H8Ousing the provided moles and MM. Assuming that we will be using 1 kg of solvent/water

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