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(a) Given Palmitic acid weight =112 g Palmitic acid Molecular weight = 256.4241 g/mol Volume of solvent(benzene) = 725 ml 1. Molarity of the solution =(Wt/M.Wt)x(1000/V in mL) = (112x1000)/(256.4241x725) = 0.602 moles/L

(a) Given Palmitic acid weight =112 g Palmitic acid Molecular weight = 256.4241 g/mol Volume of solvent(benzene) = 725 ml 1. Molarity of the solution =(Wt/M.Wt)x(1000/V in mL) = (112x1000)/(256.4241x725) = 0.602 moles/L

Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter6: Solutions And Colloids
Section: Chapter Questions
Problem 6.29P
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How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?
(a) Given Palmitic acid weight = 112 g %3D Palmitic acid Molecular weight = 256.4241 g/mol Volume of solvent(benzene) = 725 ml 1. Molarity of the solution =(Wt/M.Wt)x(1000/V in mL) = (112x1000)/(256.4241x725) = 0.602 moles/L 2. Molality of the solution = (moles of solute)/ (kilograms of solvent) Volume of solution = 725 mL Density of the solution = 0.902 g/ml Weight of the solution = Densityxvolume %3D = 0.902 x 725 =653g=0.653kg Now Molality of the solution = (112/256.4241)/0.653=0.6688 moles/kg 3. Mass percent of palmitic acid = (Wt. Of solute/Wt. Of solution)x100 %3D %3D = (112/653)x100 = 17.15% (b) Freezing point depression = Freezing point molar constant of solvent x molality of solute = 5.10 x 0.6688 = 3.14 Freezing point of solution = 5.50 - 3.14 = 2.36 °C
Transcribed Image Text:(a) Given Palmitic acid weight = 112 g %3D Palmitic acid Molecular weight = 256.4241 g/mol Volume of solvent(benzene) = 725 ml 1. Molarity of the solution =(Wt/M.Wt)x(1000/V in mL) = (112x1000)/(256.4241x725) = 0.602 moles/L 2. Molality of the solution = (moles of solute)/ (kilograms of solvent) Volume of solution = 725 mL Density of the solution = 0.902 g/ml Weight of the solution = Densityxvolume %3D = 0.902 x 725 =653g=0.653kg Now Molality of the solution = (112/256.4241)/0.653=0.6688 moles/kg 3. Mass percent of palmitic acid = (Wt. Of solute/Wt. Of solution)x100 %3D %3D = (112/653)x100 = 17.15% (b) Freezing point depression = Freezing point molar constant of solvent x molality of solute = 5.10 x 0.6688 = 3.14 Freezing point of solution = 5.50 - 3.14 = 2.36 °C
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