Wednesday
June 19, 2013

# Homework Help: Chemistry

Posted by Toni on Sunday, March 4, 2007 at 7:22pm.

Ok, in this question, the drops is throwing me off totally, can someone help! 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 in this product if 25 drops of .683M NaOH solution are required to neutralize 10 drops of cleaning vinegar? Assume that the density of the cleaning vinegar is the same as the density of regular vinegar, which is 1.005g ml^-1

drops vinegar x M = drops NaOH x M
solve for M vinegar and go from there. Post your work if you get stuck.

Is this it?
25 drops (.683M NaOH) = 10 drops Acetic Acid (X)
17.075 = 10(x)
X= 1.70 x 100% = 170% ?

It's a start.
X = 1.7075 which rounds to 1.7 to two sifnificant figures (25 and 10 drops have two s.f.). What is the unit of X? Molar.
What is molar?
Mols/Liter.
So now you know the vinegar is 1.7 mols/L. Can you change that to grams acetic acid/L? Then you know the density of the solution. Will that give you the mass of a liter of solution. And if you know the grams acetic acid and the mass of the solution, can you calculate percent?

I notice on the original post you had 0.683 and on the repost you wrote 0.0683. Make sure this one is the right one of 0.683.

Ok, here is where I get confused again.
I know moles of acetic acid = moles of vinegar so I need to convert

1.7 mol/L vinegar x 1L/1000g vin X 60g acetic acid/1 mole acetic acid (I can' set it up to cancel correct) but I eventually get 1.7 X 60 /1000 = 0.102 X 100% = 10.2% acetic acid?

The correct post is .683, did you see the post for my work?

The (X) stands for the variable I am looking for and the number should be .683M NaOH,
I get (1.7mol/L vinegar) (1L/100g vin) (60g acetic acid/1 mole acetic acid) = 1.7 x 60/1000 = 0.102 x 100% = 10.2% acetic acid?

No one has answered this question yet.

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