Suppose you titrate a 25.00 mL sample of vinegar with 17.62 mL of a standardized .1045 N solution of NaOH.
a.) What is the molarity of acetic acid in the vinegar?
b.) What is the mass of acetic acid contained in 1.000L of vinegar?
a.) What is the molarity of acetic acid in the vinegar?
b.) What is the mass of acetic acid contained in 1.000L of vinegar?
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(17.62 mL) x (0.1045 N) / (25.00 mL) = 0.07365 N = 0.07365 M acetic acid
Supposing it is the same vinegar in parts a. and b.:
(1.000 L) x (0.07365 mol/L) x (60.0522 g CH3CO2H/mol) = 4.423 g acetic acid
Supposing it is the same vinegar in parts a. and b.:
(1.000 L) x (0.07365 mol/L) x (60.0522 g CH3CO2H/mol) = 4.423 g acetic acid