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: To what volume (in mL) must a 3.45 M lead(II) nitrate solution be diluted in order to make 450.0 mL of a 0.990 M solution of lead(II) nitrate?A. 129
B. 109
C. 101
D. 56.0
E. 45.0

Problem

To what volume (in mL) must a 3.45 M lead(II) nitrate solution be diluted in order to make 450.0 mL of a 0.990 M solution of lead(II) nitrate?

A. 129
B. 109
C. 101
D. 56.0
E. 45.0

Solution

We are being asked of the volume of 3.45 M lead(II) nitrate solution be diluted in order to make 450.0 mL of a 0.990 M solution of lead(II) nitrate.

When dealing with dilution we will use the following equation:

M1V1 = M2V2

M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume


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