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Hi, could someone help me with these calculations for my lab;

Given: Volume of vinegar analyzed:5 mL
Con'c of NaOH: 0.09890M
Avg. V of NaOH from titration:43.75 mL

Moles of NaOH required to reach the equivalence point: ? Would this be (0.09890M)(0.04375L)=0.004327 moles

Moles of acetic acid 5.00mL of vinegar:? Would this be (0.0500L)(60.0g/mol)...not sure

Mass of acetic acid in the 5.00mL sample of vinegar: ? Would this be (0.004327 moles)(60.0g/mol)=0.2596g

Volume of the acetic acid in the 5.00mL sample of vinegar:? Not sure

% by volume of acetic acid in the vinegar:?

  • Chemistry -

    Yes and no.
    moles NaOH = M x L and you are correct.
    moles CH3COOH must be the same.
    grams CH3COOH = molx x molar mass.
    Thus, that is grams CH3COOH in 5 mL if you titrated a 5.00 mL sample. You want to know the g/100 mL for % w/v.
    grams in 100 mL = grams in 5 mL x 20 = ?? (because grams x (100/5) = ??)

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