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# Alcohol watch

1. The density of ethanol is 0.789g cm3 we can convert the concentration to ABV% Alcohol by volume

Concentration is 0.04

I'm not too sure what to do?

2. (Original post by gerrard1892)
The density of ethanol is 0.789g cm3 we can convert the concentration to ABV% Alcohol by volume

Concentration is 0.04

I'm not too sure what to do?

what are the units of the concentration you have been given? You need to know this in order to calculate ABV

For example, is it 0.04 g dm-3 ?
3. (Original post by Plato's Trousers)
what are the units of the concentration you have been given? You need to know this in order to calculate ABV

For example, is it 0.04 g dm-3 ?
The units are g cm3

i calculated the moles the equation n = c x v because concentration is in dm-3 so i didnt divide by 1000.

moles = 0.04 x 1 = 0.04
c x v
then it said to work out the mass of ethanol present int 1dm-3

I did 0.04 x 46 for the grams = 1.84 g

so 1.84 g in 1000 cm3
the next past was the mass in 100 cm3 this is the % by mass

so if 1.84 is 1000 cm3
0.184 must be 100 cm 3

i then did density = mass / volume re arranged it to get volume

volume = mass / density

so 0.184 / 0.789 = 0.233 im stuck with this bit am i ok up to now?
4. (Original post by gerrard1892)
The units are g cm3

i calculated the moles the equation n = c x v because concentration is in dm-3 so i didnt divide by 1000.

moles = 0.04 x 1 = 0.04
c x v
then it said to work out the mass of ethanol present int 1dm-3

I did 0.04 x 46 for the grams = 1.84 g

so 1.84 g in 1000 cm3
the next past was the mass in 100 cm3 this is the % by mass

so if 1.84 is 1000 cm3
0.184 must be 100 cm 3

i then did density = mass / volume re arranged it to get volume

volume = mass / density

so 0.184 / 0.789 = 0.233 im stuck with this bit am i ok up to now?
You don't need to worry about moles at all. Neither g cm-3 nor ABV are molar measurements.

It's simpler than that.

You have 0.04g of ethanol in every cm3. You can calculate the volume of 0.04g of ethanol from the density (0.789 g cm-3)

volume = mass / density = 0.04 / 0.789 = 0.0507 cm3

so in every 1 cm3 of solution, you have 0.0507 of alcohol.

so the ABV is (0.0507 / 1) X 100% = 5.07%
5. (Original post by Plato's Trousers)
You don't need to worry about moles at all. Neither g cm-3 nor ABV are molar measurements.

It's simpler than that.

You have 0.04g of ethanol in every cm3. You can calculate the volume of 0.04g of ethanol from the density (0.789 g cm-3)

volume = mass / density = 0.04 / 0.789 = 0.0507 cm3

so in every 1 cm3 of solution, you have 0.0507 of alcohol.

so the ABV is (0.0507 / 1) X 100% = 5.07%
Thank you, sometimes when you go down the wrong path its hard to see where your going wrong. I understand it know

+REP
6. (Original post by Plato's Trousers)
You don't need to worry about moles at all. Neither g cm-3 nor ABV are molar measurements.

It's simpler than that.

You have 0.04g of ethanol in every cm3. You can calculate the volume of 0.04g of ethanol from the density (0.789 g cm-3)

volume = mass / density = 0.04 / 0.789 = 0.0507 cm3

so in every 1 cm3 of solution, you have 0.0507 of alcohol.

so the ABV is (0.0507 / 1) X 100% = 5.07%
Ive just looked again the units of concentration are mol dm-3 rather than mol cm3
7. (Original post by gerrard1892)
Ive just looked again the units of concentration are mol dm-3 rather than mol cm3
But you didn't say it was mol cm-3. You said it was g cm-3.

So it is actually mol dm-3

Right, so what will your calculation be now?
8. (Original post by Plato's Trousers)
But you didn't say it was mol cm-3. You said it was g cm-3.

So it is actually mol dm-3

Right, so what will your calculation be now?
The concentration is in mol dm-3
The density is 0.789 g cm3

I don't know, I was going down the route that i first went down, due to it being in dm-3
9. bump
10. (Original post by Plato's Trousers)
But you didn't say it was mol cm-3. You said it was g cm-3.

So it is actually mol dm-3

Right, so what will your calculation be now?
I've looked again would i have to change the mol dm-3 to mol cm3?

Sorry for bothering you just need to get it sorted today
11. (Original post by gerrard1892)
The concentration is in mol dm-3
The density is 0.789 g cm3

I don't know, I was going down the route that i first went down, due to it being in dm-3
well, convert mol dm-3 to g dm-3 (multiply by mr)

now you know how many grams of ethanol are in a dm3

convert the grams of ethanol to a volume (divide by density)

then it's just the percentage again.

The last two steps are as I have already shown you. Only the first step is different.

So post what you get
12. (Original post by Plato's Trousers)
well, convert mol dm-3 to g dm-3 (multiply by mr)

now you know how many grams of ethanol are in a dm3

convert the grams of ethanol to a volume (divide by density)

then it's just the percentage again.

The last two steps are as I have already shown you. Only the first step is different.

So post what you get
so concentration = 0.04 mol dm-3

0.04 x 46 =1.86g

1.86/0.789 = 2.36

is that then 2.36 % ?
13. (Original post by gerrard1892)
so concentration = 0.04 mol dm-3

0.04 x 46 =1.86g

1.86/0.789 = 2.36

is that then 2.36 % ?
you have calculated that 0.04M is 1.86 g dm-3

which is 2.36 cm3 in a dm3

which is 2.36 cm3 in 1000 cm3

which is (2.36 / 1000) X 100% = 0.236% ABV
14. (Original post by Plato's Trousers)
you have calculated that 0.04M is 1.86 g dm-3

which is 2.36 cm3 in a dm3

which is 2.36 cm3 in 1000 cm3

which is (2.36 / 1000) X 100% = 0.236% ABV
Yeah thats what i got, but im suppose to have something between 1.5 and 8.5 % ABV? i dont know
15. (Original post by gerrard1892)
Yeah thats what i got, but im suppose to have something between 1.5 and 8.5 % ABV? i dont know
then your mark scheme is wrong.

I assure you 0.04M ethanol is 0.236% ABV
16. (Original post by Plato's Trousers)
then your mark scheme is wrong.

I assure you 0.04M ethanol is 0.236% ABV
Thank you, for trying to sort it out i'll juts have to write it up as 0.236% and put it in the <1.5 % tax bracket
17. (Original post by Plato's Trousers)
then your mark scheme is wrong.

I assure you 0.04M ethanol is 0.236% ABV
Can i ask you a question, I've just though of something, I've taken this by using different volumes of ethanol, we had to find the %ABV of an unknown concentration of alcohol. I used 10 ml, 8ml, 6ml, 4ml and 2ml of ethanol as my known and then used both my eye and a colorimeter to test to see if the absorbance levels were the same for a known concentration and the unknown
mine came out that the unknown has the volume of 4 ml and therefore 0.04 mol dm-3

If the definition of %ABV is in 100 ml of alcohol and mine was in 4 ml. If i times the 4ml by 25 to get 100ml Would it make sense to times the %ABV by 25 as well, to get 5.9% ABV?
18. There's not enough information here.

(Original post by gerrard1892)
I used 10 ml, 8ml, 6ml, 4ml and 2ml of ethanol as my known and then used both my eye and a colorimeter to test to see if the absorbance levels were the same for a known concentration and the unknown
These are not concentrations, they are volumes.

(Original post by gerrard1892)
mine came out that the unknown has the volume of 4 ml and therefore 0.04 mol dm-3
this doesn't make sense. How are you getting from 4 ml to 0.04M ?

(Original post by gerrard1892)
If the definition of %ABV is in 100 ml of alcohol and mine was in 4 ml. If i times the 4ml by 25 to get 100ml Would it make sense to times the %ABV by 25 as well, to get 5.9% ABV?
eh?

You seem quite confused, but you aren't giving me enough to help you. If this was a practical, can you post the instructions/method?
19. (Original post by Plato's Trousers)
There's not enough information here.

These are not concentrations, they are volumes.

this doesn't make sense. How are you getting from 4 ml to 0.04M ?

eh?

You seem quite confused, but you aren't giving me enough to help you. If this was a practical, can you post the instructions/method?
ok, I was provided with a sample of alcohol of a unknown concentration which responds to the levels of alcohol found in cider.

method -
1) make up 5 standard solutions of ethanol 10 ml, 8 ml, 6ml, 4ml, and 2ml of 1M ethanol. Then using 4 ml of the unknown sample. it says concentration in mol dm-3 = ml of ethanol used /100

2)pipette 1 ml of standard solution containing the most concentrated ethanol which is the 10ml into a test tube and add 8 ml of dichromate solution. Do this to each volumetric flask. place in water bath 30 mins

Using the colorimetry find the absorbance of each.

The absorbance with the closed reading to the unknown is the correct concentration. This was 0.04 as the reading were nearly the same.

then it says work out the mass of ethanol in 1 dm3

then what mass would be present in 100cm3 this is the % by mass

then use the density of 0.789 g cm3 to convert % by mass into a %ABV

finally state which bracket the cider sample should be taxed in.
20. ah, so you are making up your standards by adding various volumes of 1M ethanol and making up to 100 ml.

Gothca.

So your sample most closely matched the 0.04M standard.

Then the calculation above is correct.

0.04M =0.236% ABV

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