[fchs]

I'm having trouble finding where to begin in solving this problem. Please help...

We have an Avogadro’s number of M and M candies.  We know that 91 M and Ms will fill a 100.0 mL graduated cylinder.

How big of a box would we need to hold this Avogadro’s number of M and Ms?
If we set the area of the box as the area of the 48 contiguous United States, what would the height of the box have to be in order to hold this many M and Ms?
 
You must report the height of the box in km and in mi.
 
I know Avogadro's number is 6.02*10^23

From the reference I used the area of the 48 states is 7954*10^km^2

Thanks for any help

[mike]

91 M&Ms = 100 mL
6.02x10²³ M&Ms = ?? mL

6.62x10²³mL = 6.62x10²³cm³

[fchs]

I'm still lost. I'm sorry, but I just don't know what to do. Is your solution the size of the box or the height?


Thanks for any clarification

[enahs]

mL = cm³, as mike pointed out.
First figure out how many cm³ Avogadro’s  number will fill.

Then you have the volume, which you can call the volume of a cube in cm3. You can then convert to meters cubed or km cubed, ect, just by multiplying or dividing by a factor of 10.

Volume of a cube (which you now have) = Length * width * height.

If you know the volume, and you have a preset length and width, then it is just simple division to get height. Volume/(length*width) = height (they must all be in the same units).

And length * width is area (which you already have).

[mike]

The number I wrote is the volume in mL, 1 avagadro's number of M&Ms will take up.

[fchs]

Thanks sooo much! I think I'm getting there.

[fchs]

Mike, I was wondering why it would not be 6.62x10^21?

[mike]

Mike, I was wondering why it would not be 6.62x10^21?

Show me the maths.

91 M&Ms = 100 mL

1 M&M = 100/91 = 1.1 mL (approx)

so

602000000000000000000000 M&Ms = 602000000000000000000000 x 1.1 =  662000000000000000000000 mL (approx)

[enahs]

Mike, I was wondering why it would not be 6.62x10^21?

You are forgetting your 100ml in your calculation.

I am guessing you did.

6.022x10²³ M & M / 91 = x 10²¹
But it would be 6.022x10²³ M&M * (100ml / 91 M&M) = x 10²³

And do not forget you sig figs. 91 is an exact number, you can look up Alvarado’s number to as many decimal places as you need so it is not your limiting factor, your 100.0mL is your limiting significant value.

[fchs]

I found my mistake. Thanks.

[AWK]

Note, according to SI rules, the Avogadro number is devoted to microparticles (eg protons, atoms, molecules and so on). Using it for M&Ms, students and so on is incorrect.

[sjb]

Note, according to SI rules, the Avogadro number is devoted to microparticles (eg protons, atoms, molecules and so on). Using it for M&Ms, students and so on is incorrect.

It is? Do you have a reference for that? Of course, once you get to anything like macroparticles, the amount of substance involved gets rather large, but I don't recall Avogadro's number being limited to eg atoms and ions.

[AWK]

http://physics.nist.gov/cuu/Units/mole.html

      1. The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is "mol."

      2. ]When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

[Borek]

When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

I think I am the specified group of other particles

[AWK]

When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

I think I am the specified group of other particles

Man genome seems to be insuficient