any help will be great. Thanks Volume of vinegar solution used is 10.0ml Molarit
ID: 1035870 • Letter: A
Question
any help will be great. Thanks
Volume of vinegar solution used is 10.0ml Molarity of the NoOH solution (on the bottle): 0.25M Titration Number Final buret reading (ml)9.33ml6.m2.89ml Initial buret reading (mL) 43.13ml 40m 37,35ml Volume of Nao used 33.8m 34.0ml 34.46ml mL) Moles of NaOHl d 8.45milli 8.Smilli 8.615mil moles moles moles Moles of HC H,O, used0.00845ml0.0085mol0.008615mol Molanity of HC HO(M0.0845M0.085M0.08615M Average concentration (AD: 00852M Questions Calculate and record the number of moles of NaOH used for the two titrations within 0.3 mL of each other and record in the table. 1. moles NaOH-Vn (in L) xM 2. Calculate the moles of the vinegar (HC H,O) from each of the two titrations and record in the table. (See Introduction) 3. Calculate the Molarity of the vinegar (HC H,0) from each of the two titrations and record in the table. (See Introduction) 4. Calculate the average molarity s. Determine the grams of HC ,H0, from the average moles The original bottle of vinegar claims that the vinegar contains 5% acetic acid by mass. Calculate the mass % usingthe data from your experiment. Compare your resalt to what is written on the bottle. 6. , IPAGE | Lab 6BExplanation / Answer
From the data in the table
1. moles NaOH used = molarity x volume = 0.250 x volume NaOH
values are correct in the table
2. moles acetic acid = moles NaOH
correct values in the table
3. molarity vinegar solution,
Trial 1, Molarity = 0.00845 moles/0.010 L = 0.845 M
Trial 2, Molarity = 0.00850 moles/0.010 L = 0.850 M
Trial 3, Molarity = 0.008615 moles/0.010 L = 0.8615 M
4. average molarity = 0.852 M
5. mass acetic acid = 0.00852 moles x 60.05 g/mol = 0.512 g
6. mass % acetic acid = 0.512 g x 100/10 g = 5.12%
density of vinegar taken 1 g/ml
mass of vinegar taken = 10 ml x 1 g/ml = 10 g
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.