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

Answer 1

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

Answer 2

molarity of standadized NaOH soution is .1012

Answer 3

what is molarity of original vinegar

Answer 4

um... whats molarity?

Answer 5

the concentration

Answer 6

molar concentration, is defined as the amount of a constituent divided by the volume of the mixture

Answer 7

so 5.6 g acetic acid / 100 g vinegar = 5.6 % is what ur looking for right? or no?

Answer 8

where did you get 5.6 from

Answer 9

i dunno x x i forget things very easily

Answer 10

well where did you get it from i need to know

Answer 11

my head

Answer 12

yeas but can you explain

Answer 13

Suppose you start with 100 g of vinegar. (You're calculating mass/mass, so this is a good place to start.) The volume V is
V = (100 g vinegar) (1 mL vinegar / 1 g vinegar) = 100 mL vinegar

The amount of acetic acid CH3COOH in it is
(100 mL vinegar) (0.9357 mol acetic acid / 1 L vinegar) = 0.09357 mol acetic acid.

The molecular weight of acetic acid is 60 g/mol:
(0.09357 mol acetic acid) (60 g/mol) = 5.6 g acetic acid

The concentration (w/w) of acetic acid is
(5.6 g acetic acid) / (100 g vinegar) = 5.6 %

Which is really strong, so you should check my work - most vinegar is about 3% acid. ........is this what you were looking for?

Answer 14

x x iprobably did something wrong in it

Answer 15

yes

Answer 16

i'll be away until you reply

Answer 17

what is molarity of original vinegar

and the wight percent (HC2H3O2 in original vinegar)

Answer 18

gtg byee

Answer 19

@Drfoxeh

Answer 20

Typically, vinegar is a 5% solution. If 1.0 liter is 1000g (this is only a somewhat close approximation), then 50. g is vinegar and 950 g is water. 50.g divided by 60.06g/mol = 0.8325 mol. Molarity = 0.8325mol/0.950L = 0.87M.

Answer 21

hi so .87 m is the morality of original solution

Answer 22

That's the molarity of vinegar.

Answer 23

and now there asking for the weight percent( HC2H302 in Original vinegar) (assuming 1.00 L);

Answer 24

@abb0t

Answer 25

you get good grades in chem

Answer 26

ARE you good at chemisty

Answer 27

Could you try to replicate what they're asking verbatim?

Answer 28

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 diluted HC2H302 is .0783
avearge molarity of diluted vinegar is 0.7865 M
the deviation di is .000350

what is molarity of original vinegar

Answer 29

that is the question

Answer 30

idk how to go but as i think since number of moles ratio between NaoH and vinegar soltuion is 1:1, there moles should be equal....
now molarity = moles of solute per liter of solution
so i need volume of vinegar solution to calculate it....

Answer 31

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

Answer 32

THE molarity of standadized NaOH soution is .1012

Answer 33

I'm sorry, I really do not know. :( I have not learned this yet.

Answer 34

oh god i cannot help sorry

Answer 35

i have no idea what your talking about...im not in chem im in bio if this was bio i could help you...im sorry i wish i could though

Answer 36

0_0 I have no idea what to do, I'm sorry I can't help you