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 0.200 M K2SO4 solution is produced by ____.A) dilution of 250.0 mL of 1.00 M K2SO4  to 1.00 LB) dissolving 43.6 g of K2SO4 in water and diluting to a total volume of 250.0 mLC) diluting 20.0 mL of 5.00 M K2SOsolution to 500.0 mLD) dissolving 20.2 g of K2SO4 in water and diluting to 250.0 mL, then diluting 25.0 mL of this solution to a total volume of 500.0 mLE) dilution of 1.00 mL of 250 M K2SO4 to 1.00 L

Problem

A 0.200 M K2SO4 solution is produced by ____.

A) dilution of 250.0 mL of 1.00 M K2SO4  to 1.00 L

B) dissolving 43.6 g of K2SO4 in water and diluting to a total volume of 250.0 mL

C) diluting 20.0 mL of 5.00 M K2SOsolution to 500.0 mL

D) dissolving 20.2 g of K2SO4 in water and diluting to 250.0 mL, then diluting 25.0 mL of this solution to a total volume of 500.0 mL

E) dilution of 1.00 mL of 250 M K2SO4 to 1.00 L

Solution

We are asked which of the following options would yield 0.200 M K2SO4 solution


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


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