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?

Your setup and calculations are almost correct, but there is a minor error in the conversion factor. The conversion factor for the density should be 1L/1005g vin (as the given density is 1.005g/ml). So, the correct calculation will be:

(1.7 mol/L vinegar) (1L/1005g vin) (60g acetic acid/1 mole acetic acid) = 1.7 x 60/1005 = 0.10174 x 100% = 10.174% acetic acid

So, the mass percent of acetic acid in the cleaning vinegar is approximately 10.174%.

Well, well, well, looks like we have a vinegary situation here! Let me break it down for you with a splash of humor.

First, let's agree that calling it "cleaning vinegar" is a bit misleading. I mean, who wouldn't want to drink something with such a sparkling name?

Anyway, back to the drops and calculations. You're on the right track with your setup. 10 drops of cleaning vinegar (not for human consumption, remember) require 25 drops of 0.683M NaOH solution. That's like a vinegar battle royale!

Now, let's convert those drops to mols. The molar ratio is 1:1 between acetic acid and vinegar. So, if we have 1.7 mol/L vinegar, we also have 1.7 mol/L acetic acid. It's a vinegar party, my friend!

But we want to express it in terms of grams. So, let's convert 1.7 mol/L vinegar to grams. Using the density of 1.005g/mL, we can calculate that 1 liter of vinegar weighs approximately 1,005 grams. Still with me?

Now, let's convert from grams of vinegar to grams of acetic acid. Remember that epic 1:1 molar ratio? Multiply the 1.7 mol/L acetic acid by its molar mass, which is 60g/mol. That gives us 102 grams of acetic acid per liter of vinegar. That's a whole lot of acidity!

Finally, to find the mass percent, divide the mass of acetic acid (102g) by the mass of the entire solution (1,005g) and multiply by 100. Drumroll, please... the mass percent of acetic acid in the cleaning vinegar is approximately 10.2%!

So, next time you see that "cleaning vinegar" bottle, remember it's not for margaritas, it's for cleaning! Stay clear and keep those drops and measurements in check.

Let's go through the calculations step-by-step to find the mass percent of acetic acid in the cleaning vinegar.

Given:
- 25 drops of 0.683M NaOH solution are required to neutralize 10 drops of cleaning vinegar.
- Assume the density of cleaning vinegar is the same as regular vinegar, which is 1.005g/mL.

Step 1: Calculate the molarity of the cleaning vinegar.
Using the equation: drops vinegar x M = drops NaOH x M
25 drops (.683M NaOH) = 10 drops Acetic Acid (X)
17.075 = 10(x)
x = 17.075 / 10 = 1.7075

Step 2: Convert the molarity of vinegar to grams of acetic acid per liter.
1.7075 mol/L vinegar x (1 L/1000g vinegar) x (60g acetic acid/1 mol acetic acid)
= 1.7075 x 60 / 1000 = 0.10245 g acetic acid/L vinegar

Step 3: Calculate the mass percent of acetic acid in the vinegar.
Mass percent = (mass of acetic acid / mass of vinegar) x 100
Since the density of vinegar is 1.005g/mL, and it is assumed that 1 mL = 1 g, the mass of vinegar is equal to the volume of vinegar. So, the mass of vinegar in this case is 1 L.
Mass percent = (0.10245 g acetic acid / 1 g vinegar) x 100 = 10.245%

Therefore, the mass percent of acetic acid in the cleaning vinegar is approximately 10.245%.

I'm sorry for the confusion earlier. Let's go through it step by step to clarify.

To find the mass percent of acetic acid in the cleaning vinegar, you first need to determine the molar concentration of the acetic acid in the vinegar. This can be done using the equation:

drops vinegar x M = drops NaOH x M

In this case, you have 25 drops of 0.683 M NaOH solution that are required to neutralize 10 drops of the cleaning vinegar. So the equation becomes:

25 drops (0.683 M NaOH) = 10 drops Acetic Acid (X)

Now, you need to solve for X (the molar concentration of acetic acid in the cleaning vinegar).

25 drops x 0.683 M NaOH = 10 drops * X

17.075 = 10X

X = 17.075/10

X ≈ 1.7075 mol/L

Next, you need to convert the molar concentration of the vinegar to grams of acetic acid per liter. To do this, you can use the molar mass of acetic acid, which is approximately 60 g/mol.

1.7075 mol/L vinegar x (60 g acetic acid/1 mole acetic acid) = 102.45 g acetic acid/L vinegar

Now, you need to calculate the mass percent of acetic acid in the vinegar. To do this, you divide the mass of acetic acid by the mass of the solution (vinegar) and multiply by 100%.

Mass percent = (102.45 g acetic acid / 1000 g vinegar) * 100% = 10.245%

So the mass percent of acetic acid in the cleaning vinegar is approximately 10.245%.