Ch.4 - Chemical Quantities & Aqueous ReactionsWorksheetSee 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: A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate the molarity of glycerol in the solution.

Solution: A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate the molarity of glycerol in the solution.

Problem

A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate the molarity of glycerol in the solution.


Solution

We’re being asked to calculate the molarity (M) of glycerol (C3H8O3) in a solution. Recall that molarity is the ratio of the moles of solute and the volume of solution (in liters). In other words:



We first need to determine the number of moles of C3H8O3. Assuming that we have 100 g of solution, this means we have 50 g of C3H8O3 and 50 g H2O.


To find the moles of C3H8O3, we’ll need its molar mass. The molar mass of C3H8O3 is:

C3H8O3     3 C × 12 g/mol C = 36 g/mol

                  8 H × 1 g/mol H = 8 g/mol        

                  3 O × 61 g/mol O = 48 g/mol     

  Sum = 92 g/mol

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