Help me with the rest p Volume of diluted vinegar sampleused in each titration (mL)Initial buret reading (mL)Final buret reading (mL)Volume of NaOH5Volume original vinegar solution 10.00 mLVolume diluted vinegar solution 50.00ml2Dilution FactorMolarity of NaOH1/4.72x103Moles of NaOHergMoles of acetic acid (diluted)Molarity of acetic acid (diluted)Molarity of acetic acid (original)Percent by mass % (m/m)Average % (m/m)VALUECalculations:TRIAL 1TRIAL 26. 16.1TRIAL 310.00m/10.00ml 10.006.113.013.06.920.07.02.88x107¹ 3.26×10 4 3.30x10 +2.88x10 " 3.26x10" 3.30x104

Given: Volume of stock solution (V1) = 10.0 mL Volume of diluted solution (V2) = 50.0 mL molarity of NaOH = 4.72×10-2 MChemical reaction between acetic acid and NaOH is given by, CH3COOH+NaOH→CH3COONa+H2OCalculation for molarity of diluted acetic acid in trial-1: molarity=molesVolume of solutionmolarity of acetic acid=2.88×10-4 mol50.0×10-3 Lmolarity of acetic acid=0.00576 M Calculation for molarity of diluted acetic acid in trial-2: molarity=molesVolume of solutionmolarity of acetic acid=3.26×10-4 mol50.0×10-3 Lmolarity of acetic acid=0.00652 M Calculation for molarity of diluted acetic acid in trial-3: molarity=molesVolume of solutionmolarity of acetic acid=3.30×10-4 mol50.0×10-3 Lmolarity of acetic acid=0.00660 MVolume of stock solution (V1) = 10.0 mL Volume of diluted solution (V2) = 50.0 mL Calculation for molarity of original acetic acid in trial-1: Molarity of diluted solution (M2) = 0.00576 Molarity of stock solution (M1) = ? M1V1=M2V2M1=M2V2V1molarity of original acetic acid(M1)=0.00576 M×50.0 mL10.0 mLmolarity of original acetic acid(M1)=0.0288 M Calculation for molarity of original acetic acid in trial-2: Molarity of diluted solution (M2) = 0.00652 M Molarity of stock solution (M1) = ? M1V1=M2V2M1=M2V2V1molarity of original acetic acid(M1)=0.00652 M×50.0 mL10.0 mLmolarity of original acetic acid(M1)=0.0326 M Calculation for molarity of original acetic acid in trial-3: Molarity of diluted solution (M2) = 0.00660 M Molarity of stock solution (M1) = ? M1V1=M2V2M1=M2V2V1molarity of original acetic acid(M1)=0.00660 M×50.0 mL10.0 mLmolarity of original acetic acid(M1)=0.0330 MLet the density of acetic acid solution be "1.0 g/mL" as vinegar is a mixture of acetic acid and water: mass of vinegar solution=volume of solution×density of solutionmass of vinegar solution=10.0 mL×1.0 g/mLmass of vinegar solution=10.0 g Calculation for % (m/m) of acetic acid in the solution in trial-1: moles of acetic acid=molarity×volumemoles of acetic acid=0.0288 mol/L×10.0×10-3 Lmoles of acetic acid=2.88×10-4 molmass of vinegar=2.88×10-4 mol×60.05 g/molmass of vinegar=0.0173 g%(m/m) acetic acid=mass of acetic acidmass of solution×100%(m/m) acetic acid=0.017310×100%(m/m) acetic acid=0.173 % Calculation for % (m/m) of acetic acid in the solution in trial-2: moles of acetic acid=molarity×volumemoles of acetic acid=0.0326 mol/L×10.0×10-3 Lmoles of acetic acid=3.26×10-4 molmass of vinegar=3.26×10-4 mol×60.05 g/molmass of vinegar=0.0196 g%(m/m) acetic acid=mass of acetic acidmass of solution×100%(m/m) acetic acid=0.019610×100%(m/m) acetic acid=0.196 % Calculation for % (m/m) of acetic acid in the solution in trial-3: moles of acetic acid=molarity×volumemoles of acetic acid=0.0330 mol/L×10.0×10-3 Lmoles of acetic acid=3.3×10-4 molmass of vinegar=3.3×10-4 mol×60.05 g/molmass of vinegar=0.0198 g%(m/m) acetic acid=mass of acetic acidmass of solution×100%(m/m) acetic acid=0.019810×100%(m/m) acetic acid=0.198 %Calculation for average % (m/m) of acetic acid in the solution: average % (m/m) of acetic acid in the solution=0.173+0.196+0.1983average % (m/m) of acetic acid in the solution=0.189 %

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Volume original vinegar solution Volume diluted vinegar solution Dilution Factor Molarity of NaOH Volume of diluted vinegar sample used in each titration (mL) Initial buret reading (mL) Final buret reading (mL) Volume of NaOH Moles of NaOH Moles of acetic acid (diluted) Molarity of acetic acid (diluted) Molarity of acetic acid (original) Percent by mass % (m/m) Average % (m/m) Calculations: VALUE 10.00ML 50.00ml 14.78x 103 TRIAL 1 TRIAL 2 6.1 6.1 10.00m/10.00 10.00m² 13.0 20.0 7.0 2.88×10¹ 3.26x1043.30x104 2.88x10 + 3.26x10" 3.30x104 TRIAL 3 6.1 13.0 6.9

Volume original vinegar solution Volume diluted vinegar solution Dilution Factor Molarity of NaOH Volume of diluted vinegar sample used in each titration (mL) Initial buret reading (mL) Final buret reading (mL) Volume of NaOH Moles of NaOH Moles of acetic acid (diluted) Molarity of acetic acid (diluted) Molarity of acetic acid (original) Percent by mass % (m/m) Average % (m/m) Calculations: VALUE 10.00ML 50.00ml 14.78x 103 TRIAL 1 TRIAL 2 6.1 6.1 10.00m/10.00 10.00m² 13.0 20.0 7.0 2.88×10¹ 3.26x1043.30x104 2.88x10 + 3.26x10" 3.30x104 TRIAL 3 6.1 13.0 6.9

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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Question

Help me with the rest p

Volume of diluted vinegar sample
used in each titration (mL)
Initial buret reading (mL)
Final buret reading (mL)
Volume of NaOH
5
Volume original vinegar solution 10.00 mL
Volume diluted vinegar solution 50.00ml
2
Dilution Factor
Molarity of NaOH
1/4.72x103
Moles of NaOH
erg
Moles of acetic acid (diluted)
Molarity of acetic acid (diluted)
Molarity of acetic acid (original)
Percent by mass % (m/m)
Average % (m/m)
VALUE
Calculations:
TRIAL 1
TRIAL 2
6. 1
6.1
TRIAL 3
10.00m/10.00ml 10.00
6.1
13.0
13.0
6.9
20.0
7.0
2.88x107¹ 3.26×10 4 3.30x10 +
2.88x10 " 3.26x10" 3.30x104

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