Number of moles of acetic acid in the 25.00 mL sample of vinegar?

Question

molarity=M=moles solute/liter solution

Average volume=0.02481 L (24.81 mL)

Molarity Acetic acid=0.099 (0.010)

Average Moles NaOH=248.1 moles

Molarity of NaOH=.1000 M

I'm using NaOH (Sodium Hydroxide) and CH3CO2H, which is vinegar solution.

So my main question is, given all the info here, how can I figure out the number of moles of acetic acid in the 25.00 mL sample vinegar.

Please help me out! Thx much! :)

Answer 1

You should be able to see the answer right away if you look at the information given. You already have the molarity of acetic acid, it 0.099 moles/liter and you also have the average volume of acetic acid, 0.02481 liters. So simply multiply the average volume of acetic acid by the molarity of acetic acid and you'll notice that the units of volume cancel out and what you're left with is moles of acetic acid. so

(0.099mol/L)*(0.02481L) = 0.00245619 moles of acetic acid.

You can follow a similar procedure to find the moles of NaOH. As you indicated above, molarity relates moles of solute per liter of solution, so if you know the molarity of a solution you can find the number of moles present in any given volume of that solution.