Molarity problem:
Assuming the density of vinegar is 1.0 g/mL, what is the molarity of vinegar? (Use the percent by mass of acetic acid in vinegar from question 1 which is 5.3%). The molar mass of acetic acid is 60.05 g/mol.

Question

Molarity problem:
Assuming the density of vinegar is 1.0 g/mL, what is the molarity of vinegar? (Use the percent by mass of acetic acid in vinegar from question 1 which is 5.3%). The molar mass of acetic acid is 60.05 g/mol.

anonymous · Answer

Hi there,

What do you have so far?

anonymous · Answer

all i know is how to convert things and the equations
 molarity= mols of solute/liter of solution and
percent mass= solute/solution
so i think i have to use those equations for full credit on my homework but i don't know where to begin with the problem...

anonymous · Answer

Okay good - so you want to get your moles (of acetic acid) / liter of solution right?

Here are some guidelines to get you through chem - after you've figured our what you need to find, always start with your given:

$$ 1.0 grams Vinegar\over 1 mL Vinegar$$
Take a look at your other information. What else are you given? These conversions / tools will get you to your answer. 5.3% of Vinegar is Acetic Acid, in other words, there are:

$$ 5.3 grams Acetic Acid \over 100 grams Vinegar$$ 
And you also have
$$ {60.5 grams Acetic Acid \over 1 mol Acetic Acid} = 1 $$


So looking back at your given, you want to change the numerator to moles of acetic acid. Which one of your "tools" (in other words, the two relationships above) can you use to do that? As you know, you can't in this case jump directly to moles of acetic acid - try to convert your given to some unit of acetic acid first.

anonymous · Answer

@kma230 so would you convert the 60.5 grams Acetic Acid to moles solute? or the percent? or do something else?

anonymous · Answer

Do you have up to here already? You've converted your grams of Vinegar over to grams of Acetic Acid.
$${1.0gramsVinegar \over 1mLVinegar}*{5.3gramsAceticAcid \over 100gramsVinegar}$$
Then, as you mentioned, you would use the 60.5 grams per mol and change your numerator over to moles of acetic acid.
$${1.0gramsVinegar \over 1mLVinegar}*{5.3gramsAceticAcid \over 100gramsVinegar}*{1 mol Acetic Acid \over 60.5 grams Acetic Acid}$$
Cancel out your units, and you have mols of Acetic Acid over mL of Vinegar - almost at molarity! Just one final step left...can you finish it up?

anonymous · Answer

do you just have to convert mL to L so mulitply everything by 1000?

anonymous · Answer

Correct! Nice job :)

anonymous · Answer

thank you very much!! :)