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PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH3COOH (d = 1.048 g/mL), is dissolved in enough water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution?

PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH3COOH (d = 1.048 g/mL), is dissolved in enough water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH3COOH (d = 1.048 g/mL), is dissolved in enough water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution?

EXAMPLE 4-7 Calculating Molarity from Measured Quantities = 0.789 g/mL), in enough water to pro- A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d duce 250.0 mL solution. What is the molarity of ethanol in the solution? Analyze We must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculation requires the following conversions mL ethanol O density as a conversion factor and the second conversion uses mołar mass as a conversion factor. The molarity of the solution is then calculated by using equation (4.3). 2 g ethanol → mol ethanol. The first conversion uses the - Solve The number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line. 1 mol CH3CH2OH 46.07 g CH3CH,OH 0.789 g CH3CH,OH ? mol CH3CH2OH 25.0 mL CH3CH2OH × %3D 1 mL CH3CH2OН = 0.428 mol CH3CH2OH %3D To apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L. %3D Mole of solute kg of solvent 0.428 mol CH3CH2OH molarity = 1.71 M CH;CH2OH 0.2500 L soln Assess It is important to include the units in this calculation to ensure that we obtain the correct units for the final answer. When dealing with liquid solutes, be careful to distinguish between mL solute and mL soln. PRACTICE EXAMPLE A: - A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH3COOH (d water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution? 1.048 g/mL), is dissolved in enough %3D Acetone
Transcribed Image Text:EXAMPLE 4-7 Calculating Molarity from Measured Quantities = 0.789 g/mL), in enough water to pro- A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d duce 250.0 mL solution. What is the molarity of ethanol in the solution? Analyze We must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculation requires the following conversions mL ethanol O density as a conversion factor and the second conversion uses mołar mass as a conversion factor. The molarity of the solution is then calculated by using equation (4.3). 2 g ethanol → mol ethanol. The first conversion uses the - Solve The number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line. 1 mol CH3CH2OH 46.07 g CH3CH,OH 0.789 g CH3CH,OH ? mol CH3CH2OH 25.0 mL CH3CH2OH × %3D 1 mL CH3CH2OН = 0.428 mol CH3CH2OH %3D To apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L. %3D Mole of solute kg of solvent 0.428 mol CH3CH2OH molarity = 1.71 M CH;CH2OH 0.2500 L soln Assess It is important to include the units in this calculation to ensure that we obtain the correct units for the final answer. When dealing with liquid solutes, be careful to distinguish between mL solute and mL soln. PRACTICE EXAMPLE A: - A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH3COOH (d water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution? 1.048 g/mL), is dissolved in enough %3D Acetone
EXAMPLE 4-7 Calculating Molarity from Measured Quantities = 0.789 g/mL), in enough water to pro- A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d duce 250.0 mL solution. What is the molarity of ethanol in the solution? Analyze We must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculation requires the following conversions mL ethanol O density as a conversion factor and the second conversion uses molar mass as a conversion factor. The molarity of the solution is then calculated by using equation (4.3). g ethanol mol ethanol. The first conversion uses the - Solve The number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line. 0.789 g CH3CH2OH 1 mol CH;CH2OH ? mol CH3CH2OH = 25.0 mL CH3CH2OH × %3D 1 mL CHCH2CН 46.07 g CH;CH,OH = 0.428 mol CH3CH,OH To apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L. Mde of solute kq of solvent 0.428 mol CH3CH2OH molarity = 1.71 M CH3CH2OH %3D 0.2500 L soln Assess It is important to include the units in this calculation to ensure that we obtain the correct units for the final answer. When dealing with liquid solutes, be careful to distinguish between mL solute and mL soln. PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH;COOH (d water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution? = 1.048 g/mL), is dissolved in enough %3D Acetone
Transcribed Image Text:EXAMPLE 4-7 Calculating Molarity from Measured Quantities = 0.789 g/mL), in enough water to pro- A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (d duce 250.0 mL solution. What is the molarity of ethanol in the solution? Analyze We must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculation requires the following conversions mL ethanol O density as a conversion factor and the second conversion uses molar mass as a conversion factor. The molarity of the solution is then calculated by using equation (4.3). g ethanol mol ethanol. The first conversion uses the - Solve The number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line. 0.789 g CH3CH2OH 1 mol CH;CH2OH ? mol CH3CH2OH = 25.0 mL CH3CH2OH × %3D 1 mL CHCH2CН 46.07 g CH;CH,OH = 0.428 mol CH3CH,OH To apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L. Mde of solute kq of solvent 0.428 mol CH3CH2OH molarity = 1.71 M CH3CH2OH %3D 0.2500 L soln Assess It is important to include the units in this calculation to ensure that we obtain the correct units for the final answer. When dealing with liquid solutes, be careful to distinguish between mL solute and mL soln. PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enough water to produce 1.25 L of solution. What is the molarity of acetone in this solution? PRACTICE EXAMPLE B: If 15.0 mL of acetic acid, CH;COOH (d water to produce 500.0 mL of solution, then what is the molarity of acetic acid in the solution? = 1.048 g/mL), is dissolved in enough %3D Acetone
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