Molarity of NaOH= 0.1018m Data:Trial 1Trial 2M of NaOH(aq): Average M of NaOH(aq)from part IM of NaOH(aq): Average M of NaOH(aq)from part IVolume of vinegar: 10.00 mL.Volume of vinegar: 10.00 mL.Initial buret reading of NaOH (aq) : 12.35Initial buret reading of NaOH (aq): 0.00 mL.mL.Final buret reading of NaOh(aq) pink color:25.00 mL.Final buret reading of NaOh(aq) pink color:12.35 mL.Volume of NaOH(aq):Volume of NaOH(aq):Calculations:For each trial :-→ NaCH3COO(aq) + H₂O(1)NaOH(aq) + CH3COOH(aq)Moles of NaOH(aq): volume of NaOH(aq) in L. x average M of NaOH(aq) from part Ia.moles of CH3COOH : moles of CH3COOHb. Moles of NaOHM of CH3COOH = moles of CH3COOH (b) / volume of vinegar in L.Get average M of CH3COOH (aq): (M + M) / 2Mass% of CH3COOH in the vinegar solution :Molar mass of CH3COOH = 60.01 g/mole, Density of vinegar solution = 1.10 g/ moleMoles of CH3COOH (b)→g of CH3COOH : c: moles of CH3COOH xmolar mass of CH3COOH10.00 mL. vinegar x 1.10 g/mL. = 10.1 g vinegarMass% of CH3COOH = (g of CH3COOH (c) / 10.1 g vinegar) x 100Get average of mass% of CH3COOH.

Answered: Image /qna-images/answer/56a26360-8364-428e-984a-1f8b0446b2c4.jpg

Trial 1 Trial 2 M of NaOH(aq): Average M of NaOH(aq) from part I M of NaOH(aq): Average M of NaOH(aq) from part I Volume of vinegar: 10.00 mL. Volume of vinegar: 10.00 mL. Initial buret reading of NaOH (aq): 0.00 mL. Initial buret reading of NaOH (aq) : 12.35 mL. Final buret reading of NaOh(aq) pink color: 12.35 mL. Final buret reading of NaOh(aq) pink color: 25.00 mL. Volume of NaOH(aq): Volume of NaOH(aq): Calculations: For each trial : NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H₂O(1) a. Moles of NaOH(aq): volume of NaOH(aq) in L. x average M of NaOH(aq) from part I b. Moles of NaOH → moles of CH3COOH: moles of CH3COOH M of CH3COOH = moles of CH3COOH (b) / volume of vinegar in L. Get average M of CH3COOH (aq): (M+ M)/2 Mass% of CH3COOH in the vinegar solution : Molar mass of CH3COOH = 60.01 g/mole, Density of vinegar solution = 1.10 g/mole Moles of CH3COOH (b) →g of CH3COOH : c: moles of CH3COOH x molar mass of CH3COOH 10.00 mL. vinegar x 1.10 g/mL. = 10.1 g vinegar Mass% of CH3COOH = (g of CH3COOH (c) / 10.1 g vinegar) x 100 Get average of mass% of CH3COOH.

Trial 1 Trial 2 M of NaOH(aq): Average M of NaOH(aq) from part I M of NaOH(aq): Average M of NaOH(aq) from part I Volume of vinegar: 10.00 mL. Volume of vinegar: 10.00 mL. Initial buret reading of NaOH (aq): 0.00 mL. Initial buret reading of NaOH (aq) : 12.35 mL. Final buret reading of NaOh(aq) pink color: 12.35 mL. Final buret reading of NaOh(aq) pink color: 25.00 mL. Volume of NaOH(aq): Volume of NaOH(aq): Calculations: For each trial : NaOH(aq) + CH3COOH(aq) → NaCH3COO(aq) + H₂O(1) a. Moles of NaOH(aq): volume of NaOH(aq) in L. x average M of NaOH(aq) from part I b. Moles of NaOH → moles of CH3COOH: moles of CH3COOH M of CH3COOH = moles of CH3COOH (b) / volume of vinegar in L. Get average M of CH3COOH (aq): (M+ M)/2 Mass% of CH3COOH in the vinegar solution : Molar mass of CH3COOH = 60.01 g/mole, Density of vinegar solution = 1.10 g/mole Moles of CH3COOH (b) →g of CH3COOH : c: moles of CH3COOH x molar mass of CH3COOH 10.00 mL. vinegar x 1.10 g/mL. = 10.1 g vinegar Mass% of CH3COOH = (g of CH3COOH (c) / 10.1 g vinegar) x 100 Get average of mass% of CH3COOH.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter7: Statistical Data Treatment And Evaluation

Section: Chapter Questions

Problem 7.27QAP

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