Viney Trial 3 Part C. Mass Percentage Composition of Commercial Trial 2T Trial 1

Question

Viney Trial 3 Part C. Mass Percentage Composition of Commercial Trial 2T Trial 1 Standardized average molarity of NaOH(ag) (copy from Part A), M 25.00 Volume of vinegar used to prepare 10 times diluted solution in a 25.00 25.00 250-mL volumetric flask, mL 25.00 Volume of diluted solution taken for titration in each 25.00 25.00 rial, mL Final buret reading 39.8 0 39.8 of ttrant, NaOH(aq), mL 42. Initial buret reading of titrant, NaOH(aq), mL Volume of NaOH(aq) added (calculated by subtraction), mL 42. to reach the end point of titration Molarity of diluted vinegar solution calculated in each trial, M O Molanity of the original commercial vinegar, M 0100 m , 100 Number of moles of acetic acid CH COOH, in 25.00-mL sample of the original commercia vinegar, mol Mass of acetic acid, in 25.00-mL sample of the commercial vinegar CH3COOH' 9 Total mass of 25-mL sample of the commercial vinegar 25.13 25.13 25.13 vinegar Percent by mass of acetic acid in commercial vinegar calculated in each trial% , mCH COOH -100% myjnegar Average percent by mass of acetic acid in commercial vinegar (report 3 sig, fig. %

Explanation / Answer

Calculation of concentration of acetic acid in vinegar solution

Trial 1,

moles NaOH used = molarity x volume

                              = 0.1 M x 41.8 ml = 4.18 mmol

moles CH3COOH present = 4.18 mmol

molarity of vinegar solution = 4.18 mmol/25 ml = 0.1672 M

mass CH3COOH acid present = 4.18 mmol x 60.05 g/mol/1000 = 0.251 g

mass% acetic acid = (0.251 g/25.13 g) x 100 = 0.992%

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Trial 2,

moles NaOH used = molarity x volume

                              = 0.1 M x 42 ml = 4.2 mmol

moles CH3COOH present = 4.2 mmol

molarity of vinegar solution = 4.2 mmol/25 ml = 0.168 M

mass CH3COOH acid present = 4.2 mmol x 60.05 g/mol/1000 = 0.252 g

mass% acetic acid = (0.252 g/25.13 g) x 100 = 1.003%

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Trial 3,

moles NaOH used = molarity x volume

                              = 0.1 M x 39.8 ml = 3.98 mmol

moles CH3COOH present = 3.98 mmol

molarity of vinegar solution = 3.98 mmol/25 ml = 0.1592 M

mass CH3COOH acid present = 3.98 mmol x 60.05 g/mol/1000 = 0.239 g

mass% acetic acid = (0.239 g/25.13 g) x 100 = 0.951%

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average mas% CH3COOH in vinegar = 0.982%

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