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) = ??)
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:?
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