Chemistry help

[jai6638]

Hey guys..

Need some help with this question. Its more like a math question than a chem one.... Could you please verify if my solution is correct?

Q) I recently purchased a roll of aluminum foil for $1.59. The area of the roll as lisetd on the box is 6.96 square meters. I want to know specifically what I am getting for my money... Calculate the cost in dollars of a single aluminum atom in the roll of aluminum foil that I purchased today.

A) Density of Aluminum = 2700 kg/(m^(3)) = 2.7E6 g/m3

Since the thickness of the roll is probably negligible, I will take 6.96 to be the "volume of the roll" giving it the units m3.

Hence, (2.7E6)*(6.96) = 1.8792E7 g
Atomic mass of Aluminum = 26.98g
Therefore, No of moles = (1.8792e7g)/(26.98) = 6.965E5

Now, one mol of Al = 6.022E23 atoms
so, 6.965E5 mol of Al= x atoms

Therefore, X = 6.965E5 * 6.022E23 = 4.194418977E29 atoms

Cost= ($1.59)/(4.1944E29) = 3.79E-30 dollar/atom

Is that correct?

Thanks much..

 

[Heisenberg]

Originally posted by: jai6638
Since the thickness of the roll is probably negligible, I will take 6.96 to be the "volume of the roll" giving it the units m3.

Uhh...no, that would be assuming the roll is 1 m thick.

 

[tmc]

area = 6.96 m2

if you assume volume = 6.96 m3, then you are assuming that it is 1m thick, which imho is not a good assumption.

 

[jai6638]

Damn.... not sure what I was thinking... How would I figure it out then?

 

[Gibsons]

I don't think you can answer the question without knowing the thickness of the foil.

 

[Heisenberg]

Originally posted by: jai6638
Damn.... not sure what I was thinking... How would I figure it out then?

Just assume the roll is a millimeter thick. That should certainly be good to within an order of magnitude.

 

[jai6638]

I see. alright thanks... Is the procedure in my solution correct? If so, then all I have to do is just switch to mm for the thickness..

 

[jai6638]

bump

 

[gsethi]

Tell your prof that you cannot solve that problem since the thickness is not given. Instant A 😛

 

[jai6638]

lol.. he'll probably find a way out of that situation...

Anyhow, making the corrections, I got the answer to be: 3.79E-27 dollar/atom. Does this seem right?

 

[Born2bwire]

I recall having this problem in high school. I think we were to assume that the width of the foil was one atom.

 

[jai6638]

However, I need a measurement such as 1mm so that I can convert area into volume to make the units m3..

 

[Slammy1]

But even gold foil, the thinnest of foils, is 10,000 atoms think. A closer approximation would be 0.02mm = 2e-5 m.
http://en.wikipedia.org/wiki/Aluminium_foil

6.96m^2 x 2e-5m = 1.392e-4 m^3

 

[Born2bwire]

Originally posted by: jai6638
However, I need a measurement such as 1mm so that I can convert area into volume to make the units m3..

Yes, one aluminum atom.