Experiment 11: ACID-BASE TITRATIONSample Data Collection and Results PagesVolume of NaOH used in the Titration:Rough trialTrial 3Trial 2Trial 1Initial reading (mL)N1.0SMLo.00mLO .00m. OOmFinal reading (mL)43.15 mL21.50mL21.60ML21.45MLVolume dispensed (mL)2-1.55MLZI.5mL21.SOML21.0OnL21. o0mLAverage volume of NAOH used (Trials 1-3)O. 2012 MMolarity of NaOHMOLESMOLARITYAUnknown Vinegar #_LITERSCalculation of moles HC2H302 in 5.00 mL vinegar:21. 00m L NaoHMous of cetie acid

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Asked Nov 20, 2019
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With the attached data, how do I calculate the mass percent of vinegar?

Experiment 11: ACID-BASE TITRATION
Sample Data Collection and Results Pages
Volume of NaOH used in the Titration:
Rough trial
Trial 3
Trial 2
Trial 1
Initial reading (mL)
N1.0SML
o.00mL
O .00m
. OOm
Final reading (mL)
43.15 mL
21.50mL
21.60ML
21.45ML
Volume dispensed (mL)
2-1.55ML
ZI.5mL
21.SOML
21.0OnL
21. o0mL
Average volume of NAOH used (Trials 1-3)
O. 2012 M
Molarity of NaOH
MOLES
MOLARITY
A
Unknown Vinegar #_
LITERS
Calculation of moles HC2H302 in 5.00 mL vinegar:
21. 00m L NaoH
Mous of cetie acid
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Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Rough trial Trial 3 Trial 2 Trial 1 Initial reading (mL) N1.0SML o.00mL O .00m . OOm Final reading (mL) 43.15 mL 21.50mL 21.60ML 21.45ML Volume dispensed (mL) 2-1.55ML ZI.5mL 21.SOML 21.0OnL 21. o0mL Average volume of NAOH used (Trials 1-3) O. 2012 M Molarity of NaOH MOLES MOLARITY A Unknown Vinegar #_ LITERS Calculation of moles HC2H302 in 5.00 mL vinegar: 21. 00m L NaoH Mous of cetie acid

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Expert Answer

Step 1
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The veaction e Naon mith CHzc00n is as brellons: > CH2COONA + H20 NaOH CH2COOH

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Step 2
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olue NaoH 2160ML Holan ty oF NaoH = o»20|2 M MOles Lx02012 M. NAOH= 21.60ML X O-0043 md

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Step 3
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According tn the veactou, 1 mob. oNao neutaliu mole CCHCOOH Therefore o-0043 md O0013 Mo NAOH nould neutalize CH3C0OH The Moles x wolar mass CH2COOH Tams

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