Question

Molarity of standardized NaOH solution is 0.1012
final buret reading, mL IS 31.21
INITIAL buret reading, mL IS 15.72
VOLUME OF TITRANT, mL is 15.49
Moles of NaOH used is 1.568*10^-3 mol
Moles of HC2H302 neutralized by the NaOH is 1.568*10^-3 Mol
Molarity of dill HC2H302 is .0783
avearge molarity of diluted vinegar is 0.7865 M
the deviation di is .000350

what is molarity of original vinegar

The volume of diluted vinegar used in each titration is 20 ml .02 L

The Molarity of standadized NaOH soution is .1012

Answer 1

@agent0smith

Answer 2

I need to first find the molarity of original vinegar?

and then the weight percent of HC2H302 in original vinegar?

Answer 3

is this a PROJECT?

Answer 4

no its chem, it was a lab

Answer 5

I need to first find the molarity of original vinegar?

and then the weight percent of HC2H302 in original vinegar?

Answer 6

ok

Answer 7

jus a sec k

Answer 8

http://chemistry.about.com/od/dictionariesglossaries/g/defmolarity.htm

Answer 9

what was the Q ur lookin for again?

Answer 10

I need to first find the molarity of original vinegar?

and then the weight percent of HC2H302 in original vinegar?

Answer 11

that is the question

Answer 12

ok

Answer 13

tbh i havent got that far in Chemistry but i'll see what i can do ok

Answer 14

@arabpride

Answer 15

@Destinymasha @ganeshie8 @inkyvoyd

Answer 16

@triciaal @lawls

Answer 17

I need to first find the molarity of original vinegar?

and then the weight percent of HC2H302 in original vinegar?

Answer 18

sorry i lost my connect, im at work and wifi suck in the operating room

Answer 19

I already explained to you how to solve this.

Answer 20

You dont even tell us what volume of sample you diluted.

Answer 21

You are not supply all the information a person would need to solve this problem. Why should someone spend their free time bothering you for information that you should have included in your question in the first place.

Answer 22

supplying*

Answer 23

@oOKawaiiOo

Answer 24

Molarity of standardized NaOH solution is 0.1012
final buret reading, mL IS 31.21
INITIAL buret reading, mL IS 15.72
VOLUME OF TITRANT, mL is 15.49
Moles of NaOH used is 1.568*10^-3 mol
Moles of HC2H302 neutralized by the NaOH is 1.568*10^-3 Mol
Molarity of dill HC2H302 is .0783
avearge molarity of diluted vinegar is 0.7865 M
the deviation di is .000350
The volume of diluted vinegar used in each titration is 20 ml .02 L

The Molarity of standadized NaOH soution is .1012

Answer 25

I need to first find the molarity of original vinegar?

and then the weight percent of HC2H302 in original vinegar?

Answer 26

This is too complex for me. Sorry.

Answer 27

I bet @abb0t can solve this! I want to see the magic worked.

Answer 28

actuallly i did 4 experiments of the titration

Answer 29

The volume of titrant,
ml is 15.49
then it was 15.53
then it was 15.59
then it was 15.61

Answer 30

No one can solve this problem because this person has failed to provide the volume of sample (acetic acid) she diluted.

0.7865 M average molarity(20mL/Volume of sample transferred) = Molarity of Original Acetic acid solution

Then I explained how to do %weight in the previous question all very easy stuff. Not complex in the least, this person just refuses to take the time to present their question in a clear manner.