Calculate the mass % of acetic acid in the vinegar in the previous exercise (question#1 above). Assume that the density of the vinegar is 1.01 g mL.

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Chapter1: Chemical Foundations
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Give answer questions 2,3&4

 

1) Given : volume of vinegar solution used = 2.50 mL

concentration of NaOH = 0.0960 M

volume of NaOH required = 34.90 mL

The reaction between NaOH and acetic acid is given by 

NaOH + CH3COOH -------> CH3COONa + H2O

From the above reaction we can say that 1 mole of NaOH is required to neutralise 1 mole of CH3COOH and reach equivalence point.

Hence moles of NaOH required = moles of CH3COOH present

Since moles = concentration X volume of solution 

=> concentration of acetic acid X volume of vinegar solution = concentration of NaOH X volume of NaOH solution 

Hence substituting the values we get 

concentration of acetic acid X 2.50 = 0.0960 X 34.90 

=> concentration of acetic acid = 1.34 M approx

Given : 500 mL of vinegar solution = 0.5 L                                                                    (since 1 L = 1000 mL)

Since moles = concentration X volume of solution in L

=> moles of acetic acid = 1.34 X 0.5 = 0.67 mol

Since molar mass of acetic acid = Atomic mass of C X 2 + Atomic mass of H X 4 + Atomic mass of O X 2 = 12 X 2 + 4 X 1 + 16 X 2 = 60 g/mol

Since mass = moles X molar mass 

-> mass of acetic acid in 500 mL vinegar = 0.67 X 60 = 40.2 g

Calculate the mass % of acetic acid in the vinegar in the previous exercise (question#1
above). Assume that the density of the vinegar is 1.01 g mL.
Transcribed Image Text:Calculate the mass % of acetic acid in the vinegar in the previous exercise (question#1 above). Assume that the density of the vinegar is 1.01 g mL.
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