ANALYSIS OF HOUSEHOLD VINEGAR (NEW Data/Cale Page) Balanced Reaction Equation: 1

Question

ANALYSIS OF HOUSEHOLD VINEGAR (NEW Data/Cale Page) Balanced Reaction Equation: 1 C If s CcoH- I NeoH- Data: Concentration of NaOH solution, M 0.154M CH Path 1-Molar Concentration Trial 1 Trial 2 Final Vinegar Buret reading, mLI 29.20n3146m aital Vinegar Buretding. m 21.6m 29drmh e Use a buret t 2 mL of vi each fla Volume of Vinegar, m 2,05 mL 2.25 mL added by Final NaOH Buret reading, mL Initial NaOH Buret reading, mL mL Volume of NaOH added, mL! 9,210 m MolesofNOHadded.m ol ,00 143 mo .00 115 mol Moles of Acetic Acid, mol 00143 mol 00 loSmol 1.696 M |.732 centration of Acetic Acid, M Average % Acetic Acid % RAD tions: Continue on next page if needed Reagent Table Name Formula Appearan a9

Explanation / Answer

Analysis of acetic acid in vinegar solution

Trial 1 : moles of acetic acid in vinegar = 0.00143 mol

mass of acetic acid present = 0.00143 mol x 60 g/mol = 0.0858 g

Taking density of vinegar = 1.01 g/ml

mass of vinegar solution = 2.05 g x 1.01 g/ml = 2.0705 g

% acetic acid in solution = 0.0858 g x 100/2.0705 g = 41.44%

Trial 2 : moles of acetic acid in vinegar = 0.00165 mol

mass of acetic acid present = 0.00165 mol x 60 g/mol = 0.099 g

Taking density of vinegar = 1.01 g/ml

mass of vinegar solution = 2.25 g x 1.01 g/ml = 2.2725 g

% acetic acid in solution = 0.099 g x 100/2.2725 g = 43.56%

average % acetuc acid = (41.44 + 43.56)/2 = 42.5%

%RAD = sq.rt.[(42.5 - 41.44)^2 + (42.5 - 43.56)^2)/2] = 1.5%

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