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: What volume (L) of 0.18 M H2SO4 can be prepared by diluting 143 mL of 5.5 M H2SO4 ?

Solution: What volume (L) of 0.18 M H2SO4 can be prepared by diluting 143 mL of 5.5 M H2SO4 ?

Problem

What volume (L) of 0.18 M H2SO4 can be prepared by diluting 143 mL of 5.5 M H2SO4 ?


Solution

We are being asked to find the volume of 0.18 M H2SO4 can be prepared by diluting 143 mL of 5.5 M H2SO4.

When we are adding water (or solvent) to a solution to decrease its concentration, we are diluting the solution. When dealing with dilution we will use the following equation:

M1V1=M2V2

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


Calculate the final volume of the 0.18 M H2SO4 solution.

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