The question is asking the molarity of vinegar, and I am not sure if i did it right.
NaOH + HCH3COO = CH3COONa + H2O
NaOH C= 0.70mol/L
V= 14.6mL --> 0.0146L
n, NaOH = CxV
= 0.70 (0.0146)
= 0.01022mol
1 mol NaOH = 0.01022mol NaOH
----------------- -------------------------
1 molCH3COONa x mol CH3COONa
x= 1(0.01022)
x = 0.01022mol
C= n = 0.01022mol
----- ------------- = 1.02mol/L
v 0.01L
The next step is to convert the molarity to percentage by volume and I am confused on how to do that.
You should have written the problem and you didn't. You supplied data and I have tried to reconstruct the problem. I don't know if I've done that right or not. Here is what I think the problem must have said.
Calculate the molarity of vinegar if 10 mL of vinegar were neutralized by 14.6 mL of 0.70 M NaOH. Convert to percent w/v.
Your calculation of M seems to be correct at 1.02 mols/L. The molar mass of vinegar (acetic acid or CH3COOH is 60 grams/mol).
1.02 mols/L x 60 grams/mol = about 60 grams CH3COOH/1000 mL solution.
% w/v = grams solute/100 mL. Your 60 g/1000 mL is 6 g/100 = about 6% w/v. You will need to redo this last part since I've estimated. I usually see vinegar on the store shelves as 4% or 5% so the about 6% sounds a little hight to me.
Sorry for not stating the question clearly, this was a lab report.
The lab was finding the concentration of vinegar. Maybe this will give you more of the idea what it is about.
This was the trial data we did: Trail 1 Trail 2 Trail 3 Trail 4 Trail 5
Concentration of base(NaOH) solution (M) 0.70 0.70 0.70 0.70 0.70
Initial reading (mL) 0mL 14.8 0 14.3 29.3
Final reading (mL) 14.8 29.5 14.3 29.3 43.7
Volume of soduim Hydroxide (mL) 14.8 14.7 14.3 15 14.4
Volume of acid (Vinegar) solution(mL) 10.0 10.0 10.0 10.0 10.0
The average volume of NaOH = 14.64 ...14.6
Balanced equation: NaOH + HCH3COO ----> CH3COONa + H2O
1) The next question was to calculate the molarity of the vinegar. I got 1.02mol/L
2) Convert the molarity above to percentage by volume (v/v%)?
Sorry, for the confusion. The question was not asking for the weight per volume.
Your 1.02 M looks ok to me. However, your last sentence in your first post said "The next step is to convert the molarity to percentage by volume and I am confused on how to do that." and I've shown you how to do that. 1.02 mols/L x 60 g/mol = about 60 g/1000 mL which is about 6 g/100 mL and that is the definition of percent by volume so your sample is about 6% w/v.
Oh, I was confused on where you got the 60g from. Did you find the molar mass of vinegar and multiply it by the number of moles to get around 60.07g .....60g. Also the question gave the density of the acetic acid 1.05g/mL, am I suppose to incorporate that with anything?
Yes, the 60 g is the molar mass of CH3COOH. If you want the percent weight/volume then you don't need the density. You would need that if you want percent weight/weight.
Sir, I think you are getting the percentage weight/volume and the volume/volume mixed up.
As far as I know shouldn't it be set up more like this..
v/v % = volume of acetic acid divided by volume of vinegar × 100
And the density is 1.05g/mL
Well, first you are right that I have confused v/v with w/v but when I read the problem I misconstrued the meaning because w/v is the usual question. I just didn't read the problem closely enough.
Also, I read the density of the SOLUTION as 1.05 g/mL and the problem may have meant the density of pure acetic acid is 1.05 g/mL.That wasn't clear in your post. I looked up the density of glacial acetic acid and it gave 1.05 g/mL; therefore, if you have about 0.6 g acetic acid in that 10 mL, the volume of the acetic acid is v = m/d = about 0.6/1.05 = ?, then
%v/v = (volume acetic acid/volume solution)*100 or (? mL/10 mL)*100 = xx
When I did the equation I got 0.0583 as the percentage and I am sure I did it wrong.
This is what I did...
First I calculated the molar mass of HCH3COO = 60.06g/mol
Then I found the moles by C X V = 1.02mol/l (0.01L) = 0.0102mol
mass of acetic acid--> n x M
= 0.0102 (60.06)
= 0.6126g
Volume of acetic mass= mass acid divided by density acetic aid
=0.6126 divided by 1.05
= 0.583mL---> 0.00583L
v/v% = 0.000583 divided by 0.01L x 100 = 0.0583
Can you please check my claculation? Also my teacher left this info to guide us.
1.05 mol acid
------------------- --> ----- -------- =L acid
1L L L L solid
1.05mol--->mass ---> volume
Il rewrite the last part in case it wasn't clear 1.05mol acid/1L --> ?/L ?/L ---> Lacid/Lsolid
1.05 mol --> mass--> volume
OK. First, you calculated the vinegar in your first post as 1.02 M and that is correct. The last post is confusing how you did that so let me start over. I will use 0.0146 L for the volume of NaOH, 0.01 L for the volume of vinegar used, 0.70 M as the molarity of the NaOH and 60.1 as the molar mass of CH3COOH.
CH3COOH + NaOH ==> CH3COONa + H2O
mols NaOH = c*v = 0.70 x 0.0146 = 0.0102 mols.
mols acid = mols NaOH (because the equation shows 1 mol NaOH = 1 mols acid).
Then M acid in the vinegar = mols/L = 0.0102/0.01 = 1.02 M which is your answer in your first post.
So you have 1.02 mols acid in 1L of 1.02 M acid and grams = mols x molar mass = 1.02 x 60.1 = 61.3 grams CH3COOH in 1L, The volume of CH3COOH then is volume= mass/density = 61.3 g/1.05 g/mL = 58.4 mL in that 1000 mLvinegar. The definition of % v/v is (volume solute/volume solution)*100 = (58.4 mL acid/1000 mLsolution)*100 =5.84% v/v
You may want to use 60.06 so your numbers will change a little. I think you converted mols in 1 L to mols in the 10 mL and that's ok but I find that the more often I change units the more mistakes I make so I choose to keep everything in 1 L and work in mols/L. I hope this helps. I think I see your error. I will copy the last part of your work, then bold what you should have done.
"First I calculated the molar mass of HCH3COO = 60.06g/mol
Then I found the moles by C X V = 1.02mol/l (0.01L) = 0.0102mol(<.b>this is 0.0102 mols in 10 mL sample).
mass of acetic acid--> n x M
= 0.0102 (60.06)
= 0.6126g
Volume of acetic mass= mass acid divided by density acetic aid
=0.6126 divided by 1.05
= 0.583mL---> 0.00583L
0.583 mL in the 10 mL sample or 5.83 mL in 100 which is 5.83% v/v or 58.3 mL in 1L (not 0.00583) but that is still 5.83 mL/100 mL which is 5.83%.
v/v% = 0.000583 divided by 0.01L x 100 = 0.0583"