Post lab.question) Experiment result: Concentration of HC2H3O2 = 0.00855 mole HC

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Post lab.question)
Experiment result: Concentration of HC2H3O2 = 0.00855 mole HC2H3O2/ 0.0100L(10mL) solution =0.855M The manufacturer's vinegar used in thisexperiment claims that the vinegar contains 5% acetic acid byweight. Use your results and a density of 1.0 g/mL to investigatethis claim. Post lab.question)
Experiment result: Concentration of HC2H3O2 = 0.00855 mole HC2H3O2/ 0.0100L(10mL) solution =0.855M The manufacturer's vinegar used in thisexperiment claims that the vinegar contains 5% acetic acid byweight. Use your results and a density of 1.0 g/mL to investigatethis claim.

Explanation / Answer

molecular mass : 60.05 g/mol mass = 0.513 g in 10 mL mass in 1 L = 0.513 X 100= 51.3 g because water density is 1g/1mL, in 1 L, you have 1000g ofwater (51.3g/1000g)x100%= 5.13%

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