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: Which of the following would be used to make a 0.40 M NaCl solution (NaCl mass and water volume, respectively)A. 29.2 g, 500 mL B. 58.4 g, 4.00 L C. 11.7 g, 400 mL D. 58.4 g, 1.00 L E. 5.84 g, 250 mL

Problem

Which of the following would be used to make a 0.40 M NaCl solution (NaCl mass and water volume, respectively)

A. 29.2 g, 500 mL 

B. 58.4 g, 4.00 L 

C. 11.7 g, 400 mL 

D. 58.4 g, 1.00 L 

E. 5.84 g, 250 mL

Solution

Determine which among the choices will produce 0.40 M NaCl using the formula for molarity

Recall that molarity is

Molarity = moles of soluteL of solution


Calculate each molarity produced by each choice where the molar mass of NaCl is:

NaCl             1 Na × 22.99 g/mol Na = 22.99 g/mol
                     1 Cl × 35.45 g/mol Cl = 35.45 g/mol
                               Sum = 58.44 g/mol

A. 29.2 g, 500 mL (1 mL = 1x10-3 L)

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