Chemistry
posted by Toni HELP! on .
Ok, I think I have this one but can you please check my work? A nationally known company markets a product called "cleaning vinegar" that is not designed for human consumption. What is the Mass Percent of acetic acid (HC2H3O2) in this product, if 25 drops of .683 M NaOH solution are required to neutralize 10 drops of cleaning vinegar? Assume that the density of the cleaning vinegare is the same as the density of regular vinegar, which is 1.005 g ml^1
I got:
25(.683M NaOH) = 10(M HC2H3O2)=
17.075/10 = 1.7M of HC2H3O2 then
1.7M HC2H3O2/1L vinegar(60g HC3H3O2/1 mole HC2H2O3) = 102g HC2H3O2/1L/1000g vinegar = .102(100%) = 10.2% HC2H3O2
I know this is long, but I need for Tue. been working on for 3 days now, and part of midterm prelab, we have no tutors at our University!
Not quite. almost.
1.7 M is ok.
That x 60 = 102 g/L BUT that is not 1000 g. Since the density is 1.005 g/mL, then 1000 mL has a mass of 1005 grams. Therefore, you have 102 g/1005 g and that gives a mass percent of
[102/1005]x100=??
I know that is almost the same NUMBER (10.2 vs 10.19 which rounds to 10.2) BUT dividing by 1000 is faulty and dividing by 1005 is correct. It's the method that counts (usually).
Check my thinking. Check my arithmetic.
As an after thought, note that if your instructor pays attention to sifnificant figures, then you are allowed only two for 10 drops and 25 drops have only two s.f. in them. Therefore, the answer can have no more than two and you would report 10.%. Make sure you place the decimal after the 0 as 10.% and not 10%.
Thank You, this came just in time and I will double check the #'s!
Your a lifesaver!

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