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 4-7Calculating Molarity from Measured Quantities= 0.789 g/mL), in enough water to pro-A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (dduce 250.0 mL solution. What is the molarity of ethanol in the solution?AnalyzeWe must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculationrequires the following conversions mL ethanolO density as a conversion factor and the second conversion uses mołar mass as a conversion factor. The molarityof the solution is then calculated by using equation (4.3).2g ethanol → mol ethanol. The first conversion uses the-SolveThe number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line.1 mol CH3CH2OH46.07 g CH3CH,OH0.789 g CH3CH,OH? mol CH3CH2OH25.0 mL CH3CH2OH ×%3D1 mL CH3CH2OН= 0.428 mol CH3CH2OH%3DTo apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L.%3DMole of solutekg of solvent0.428 mol CH3CH2OHmolarity= 1.71 M CH;CH2OH0.2500 L solnAssessIt is important to include the units in this calculation to ensure that we obtain the correct units forthe final answer. When dealing with liquid solutes, be careful to distinguish between mL soluteand mL soln.PRACTICE EXAMPLE A: - A 22.3 g sample of acetone (see the model here) is dissolved in enoughwater 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 (dwater 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%3DAcetone EXAMPLE 4-7Calculating Molarity from Measured Quantities= 0.789 g/mL), in enough water to pro-A solution is prepared by dissolving 25.0 mL ethanol, CH3CH2OH (dduce 250.0 mL solution. What is the molarity of ethanol in the solution?AnalyzeWe must first calculate how many moles of ethanol are in a 25.0 mL sample of pure ethanol. This calculationrequires the following conversions mL ethanolO density as a conversion factor and the second conversion uses molar mass as a conversion factor. The molarityof the solution is then calculated by using equation (4.3).g ethanolmol ethanol. The first conversion uses the-SolveThe number of moles of ethanol in a 25.0 mL sample of pure ethanol is calculated below in a single line.0.789 g CH3CH2OH1 mol CH;CH2OH? mol CH3CH2OH =25.0 mL CH3CH2OH ×%3D1 mL CHCH2CН46.07 g CH;CH,OH= 0.428 mol CH3CH,OHTo apply the definition of molarity given in expression (4.3), note that 250.0 mL = 0.2500 L.Mde of solutekq of solvent0.428 mol CH3CH2OHmolarity =1.71 M CH3CH2OH%3D0.2500 L solnAssessIt is important to include the units in this calculation to ensure that we obtain the correct units forthe final answer. When dealing with liquid solutes, be careful to distinguish between mL soluteand mL soln.PRACTICE EXAMPLE A: A 22.3 g sample of acetone (see the model here) is dissolved in enoughwater 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 (dwater 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%3DAcetone

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Data given:
Practice Example A:
Mass of acetone = 22.3 g
Volume of solution= 1.25 L
Molarity is…