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The tartaric acid, H2C4H4O6 (150.087 g/mol ) concentration in 100.0 mL wine sample was determined using 0.4050 M sodium hydroxide standard solution. In the experiment, the %w/w of tartaric acid present in the wine sample by analyzing and titrating only 25.00 mL aliquot from the 100-ml wine sample. The endpoint of titration was reached upon the addition of exactly 43.56 mL NaOH solution? (Assume the density of the wine sample is 1.000 g/mL) H2C4H4O6 + 2NaOH⟶Na2C4H4O6+2H2O What is the total (%w/w) acid content in the wine sample?       1.324 %w/w     2.648 %w/w     5.296 %w/w     10.59 %w/w

The tartaric acid, H2C4H4O6 (150.087 g/mol ) concentration in 100.0 mL wine sample was determined using 0.4050 M sodium hydroxide standard solution. In the experiment, the %w/w of tartaric acid present in the wine sample by analyzing and titrating only 25.00 mL aliquot from the 100-ml wine sample. The endpoint of titration was reached upon the addition of exactly 43.56 mL NaOH solution? (Assume the density of the wine sample is 1.000 g/mL) H2C4H4O6 + 2NaOH⟶Na2C4H4O6+2H2O What is the total (%w/w) acid content in the wine sample?       1.324 %w/w     2.648 %w/w     5.296 %w/w     10.59 %w/w

Introductory Chemistry For Today
Introductory Chemistry For Today
8th Edition
ISBN:9781285644561
Author:Seager
Publisher:Seager
Chapter7: Solutions And Colloids
Section: Chapter Questions
Problem 7.34E
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  1. The tartaric acid, H2C4H4O6 (150.087 g/mol ) concentration in 100.0 mL wine sample was determined using 0.4050 M sodium hydroxide standard solution.

    In the experiment, the %w/w of tartaric acid present in the wine sample by analyzing and titrating only 25.00 mL aliquot from the 100-ml wine sample.

    The endpoint of titration was reached upon the addition of exactly 43.56 mL NaOH solution?

    (Assume the density of the wine sample is 1.000 g/mL)

    H2C4H4O6 + 2NaOH⟶Na2C4H4O6+2H2O

    What is the total (%w/w) acid content in the wine sample?

     

       

    1.324 %w/w
       

    2.648 %w/w
       

    5.296 %w/w
       

    10.59 %w/w
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