Question 15 Methylene chloride (CH₂Cl₂) is used in various industrial processes, in many different industries including paint stripping, pharmaceutical manufacturing, paint remover manufacturing, and metal cleaning and degreasing. A theoretical mixture is created by dissolving 26.5318mL of Methylene chloride in 241.3701mL of Diethyl ether (C₂H5)2O). What is the Molal concentration of the solution? Take note of the following values. Density: ● Methylene chloride = 1.33 g/mL Diethyl ether = 0.713 g/mL Molar Mass: • C = 12.01 g/mol • CL-25 15 g/mol

Question 15 Methylene chloride (CH₂Cl₂) is used in various industrial processes, in many different industries including paint stripping, pharmaceutical manufacturing, paint remover manufacturing, and metal cleaning and degreasing. A theoretical mixture is created by dissolving 26.5318mL of Methylene chloride in 241.3701mL of Diethyl ether (C₂H5)2O). What is the Molal concentration of the solution? Take note of the following values. Density: ● Methylene chloride = 1.33 g/mL Diethyl ether = 0.713 g/mL Molar Mass: • C = 12.01 g/mol • CL-25 15 g/mol

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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Sample Exercise 4: Calculation of molality and Molarity A solution is made by dissolving 4.35 g glucose (C6H₁2O6) in 25.0 mL of water at 25 °C. Calculate the (a) molality and (b) the Molarity of glucose in the solution. Water has a density of 1.00 g/mL. The glucose solution has a density of 1.046 g/mL. Solution (a) Calculating molality requires that we know the mass of both components separately. We then use the MM of the solute to get the moles of solute. mass H₂O: = 25.0 mL x 1.00 g 25.0 mL x mL moles glucose: 1.00 g = 25.0 g H₂O MM C6H₁₂O6 = 180.16 g/mol 4.35 g 180.16 g mol-¹ = 0.024145 mol C6H₁2 06 molality glucose: 0.024145 mol = 0.966 m 0.0250 kg H₂O Molarity glucose: (b) To get the Molarity of the glucose in the solution, we need to know the mass of the solution, then use the solution density to find the volume of the solution. mass of solution: mass H₂O + mass glucose = 25.0 g + 4.35 g = 29.35 g volume of solution: 29.35 g x 1 mL = 28.06 mL of sol'n 1.046 g = 0.860 M 0.024145…