Based from the Molarity of the palmitic acid solution that you were able to solve earlier, how will you prepare 250 mL of the said molar concentration, if you have an available 2M stock solution of Palmitic acid? How many mL of the 2M palmitic acid stock solution will you need? Final answer must be rounded off to 2 decimal places, and shall NOT have any units. (a) Given Palmitic acid weight = 112 g%3DPalmitic acid Molecular weight = 256.4241 g/molVolume of solvent(benzene) = 725 ml1. Molarity of the solution =(Wt/M.Wt)x(1000/V in mL)= (112x1000)/(256.4241x725)= 0.602 moles/L2. Molality of the solution = (moles of solute)/ (kilograms of solvent)Volume of solution = 725 mLDensity of the solution = 0.902 g/mlWeight of the solution = Densityxvolume%3D= 0.902 x 725 =653g=0.653kgNow Molality of the solution = (112/256.4241)/0.653=0.6688 moles/kg3. 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= 3.14Freezing point of solution = 5.50 - 3.14 = 2.36 °C

Given, Stock solution : Molarity M1: =2 M Volume = V1 (to calculate) Diluted solution : Molarity…

Answered: Based from the Molarity of the palmitic…

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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

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Based from the Molarity of the palmitic acid solution that you were able to solve earlier, how will you prepare 250 mL of the said molar concentration, if you have an available 2M stock solution of Palmitic acid? How many mL of the 2M palmitic acid stock solution will you need? Final answer must be rounded off to 2 decimal places, and shall NOT have any units.

(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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