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3. The manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Using y
Date Concentration of Acetic Acid in Vinegar: Results Instructor Name Partners Section Data molarity of the NaOH solution, M
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Answer #1

The average millimoles of NaOH = 0.25*(32+33+32)/3 = 8.08 mmol

The average millimoles of CH3COOH = 100*0.808 = 80.8 mmol

i.e. The moles of CH3COOH actually consumed in the reaction with NaOH = 8.08 mmol

Note: NaOH is a limiting reactant.

Here, the % by weight = 80.8/8.08 = 10%

Which is not matching with the manufacturer's claim (5%).

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