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March 6, 2015

Homework Help: Chemistry

Posted by Adam on Sunday, January 30, 2011 at 10:05pm.

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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