Domanda del libro di testo
(b) If you dilute 10.0 mL of the stock solution to a final volume of 0.250 L, what will be the concentration of the diluted solution?
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Ch.4 - Reactions in Aqueous Solution
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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TOC.tableOfContents
- Ch.1 - Introduction: Matter, Energy, and Measurement177
- Ch.2 - Atoms, Molecules, and Ions231
- Ch.3 - Chemical Reactions and Reaction Stoichiometry216
- Ch.4 - Reactions in Aqueous Solution189
- Ch.5 - Thermochemistry175
- Ch.6 - Electronic Structure of Atoms168
- Ch.7 - Periodic Properties of the Elements150
- Ch.8 - Basic Concepts of Chemical Bonding177
- Ch.9 - Molecular Geometry and Bonding Theories200
- Ch.10 - Gases179
- Ch.11 - Liquids and Intermolecular Forces113
- Ch.12 - Solids and Modern Materials154
- Ch.13 - Properties of Solutions157
- Ch.14 - Chemical Kinetics199
- Ch.15 - Chemical Equilibrium128
- Ch.16 - Acid-Base Equilibria164
- Ch.17 - Additional Aspects of Aqueous Equilibria163
- Ch.18 - Chemistry of the Environment105
- Ch.19 - Chemical Thermodynamics164
- Ch.20 - Electrochemistry171
- Ch.21 - Nuclear Chemistry122
- Ch.22 - Chemistry of the Nonmetals82
- Ch.23 - Transition Metals and Coordination Chemistry79
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry16
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 77
Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL at 25 C. Calculate the molarity of a solution of acetic acid made by dissolving 20.00 mL of glacial acetic acid at 25 C in enough water to make 250.0 mL of solution.
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Determine the mass of glacial acetic acid using its volume and density: \( \text{mass} = \text{density} \times \text{volume} \).
Calculate the number of moles of acetic acid using its molar mass: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of acetic acid (CH₃COOH) is approximately 60.05 g/mol.
Determine the volume of the solution in liters. Since the final volume of the solution is 250.0 mL, convert this to liters: \( 250.0 \text{ mL} = 0.2500 \text{ L} \).
Calculate the molarity of the acetic acid solution using the formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \).
Combine the calculated moles of acetic acid and the volume of the solution in liters to find the molarity.
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Density
Density is defined as mass per unit volume and is a crucial property of substances. In this context, the density of glacial acetic acid (1.049 g/mL) allows us to convert the volume of acetic acid used (20.00 mL) into mass, which is essential for calculating molarity.
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01:56
Density Concepts
Molarity
Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. To find the molarity of the acetic acid solution, we need to determine the number of moles of acetic acid from its mass and then divide by the total volume of the solution in liters.
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Percorso guidato
01:13Molarity
Moles and Molar Mass
Moles are a unit of measurement in chemistry that quantify the amount of substance. The molar mass of acetic acid (approximately 60.05 g/mol) is used to convert the mass of acetic acid obtained from the density calculation into moles, which is necessary for the molarity calculation.
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02:11Molar Mass Concept
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