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1. A 4.00 mL sample of vinegar (active ingredient, acetic acid) was titrated with 0.1250M NaOH according to the following equation: IL = 1000 ML HC₂H3O2 + NaOH → NaC₂H3O2+ HOH 25.19 mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 25.19 25.19 m² = = 0.025192 1000 molx vol = 0.1250 mol/ x 0. 025196 4.00_2=0.0042 = 1/7/² = 0.00314875 mol 0.0040 1000 = 0.00314875 mol = 0.787/875 MOL/L € 0.7872m 43.239 2. Calculate the grams of acetic acid in one liter of the vinegar. 0.7871875 mol/L x 60 gmol = 43.231259/2 > 64.05C Calauloto the grams of vinegar in

1. A 4.00 mL sample of vinegar (active ingredient, acetic acid) was titrated with 0.1250M NaOH according to the following equation: IL = 1000 ML HC₂H3O2 + NaOH → NaC₂H3O2+ HOH 25.19 mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 25.19 25.19 m² = = 0.025192 1000 molx vol = 0.1250 mol/ x 0. 025196 4.00_2=0.0042 = 1/7/² = 0.00314875 mol 0.0040 1000 = 0.00314875 mol = 0.787/875 MOL/L € 0.7872m 43.239 2. Calculate the grams of acetic acid in one liter of the vinegar. 0.7871875 mol/L x 60 gmol = 43.231259/2 > 64.05C Calauloto the grams of vinegar in

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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How do I calculate the grams of vinegar in one liter?
ACID-BASE TITRATION PRELAB QUESTIONS NAME Rachel MacDowell 1. A 4.00 mL sample of vinegar (active ingredient, acetic acid) was titrated with 0.1250M NaOH according to the following equation: IL = 1000 ML HC₂H3O₂ + NaOH NaC₂H3O2 + HOH 25.19 mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 25.19m²= 25.19 = 0.025192 1000 molx vol = 0.1250 mol/ x 0.02519L = 0.00314875 mol 4.00 2 = 0.0042 = 1/7/² 1000 C 0.00314875 mol 0.0040 0.7871875 mol/L € 0.7872m 2. Calculate the grams of acetic acid in one liter of the vinegar. 0.7871875 mol/L x 60 g/mol = 43.23125 3/2 3. The 4.00 mL sample of vinegar had a mass of 4.056 g. Calculate the one liter. 43.239. grams of vinegar in
Transcribed Image Text:ACID-BASE TITRATION PRELAB QUESTIONS NAME Rachel MacDowell 1. A 4.00 mL sample of vinegar (active ingredient, acetic acid) was titrated with 0.1250M NaOH according to the following equation: IL = 1000 ML HC₂H3O₂ + NaOH NaC₂H3O2 + HOH 25.19 mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 25.19m²= 25.19 = 0.025192 1000 molx vol = 0.1250 mol/ x 0.02519L = 0.00314875 mol 4.00 2 = 0.0042 = 1/7/² 1000 C 0.00314875 mol 0.0040 0.7871875 mol/L € 0.7872m 2. Calculate the grams of acetic acid in one liter of the vinegar. 0.7871875 mol/L x 60 g/mol = 43.23125 3/2 3. The 4.00 mL sample of vinegar had a mass of 4.056 g. Calculate the one liter. 43.239. grams of vinegar in
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