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# Chemistry problem with volumes - help! watch

1. Okay, so I can't figure out how to do this, and thought someone on here might be able to help

So, basically, a certain dilution of a substance contains 6 x 10^-37 molecules/dm^3

If I wanted to figure out the radius of the sphere whose volume is equal to that which allows (on average) one molecule of this substance to be present, how would I go about it?

Many thanks!
2. I'm not entirely sure, but wouldn't it depend on the density of the substance? as density=mass/volume?
3. Ooh interesting, although that is a very low conc.
First the volume for a sphere = 4pi/3 r ^3
You would need the volume to be 1.67 x 10^36 dm^3 for there to be one mole there which is 1.67 x 10^33 m^3 (that is riduclousy large about a million x the size of the sun)
So the radius of a sphere for a given volume is given by (3/4pi V)^1/3 which would make the radius about 7.36 x 10 ^10 m or 73600000 km (about the distance to mars)
This questions seems a non-sequitir are you sure the conc is correct?
4. I'm not sure. It's being diluted in water, density of which is 1kg per litre, I think.
5. (Original post by joe3469)
Ooh interesting, although that is a very low conc.
First the volume for a sphere = 4pi/3 r ^3
You would need the volume to be 1.67 x 10^36 dm^3 for there to be one mole there which is 1.67 x 10^33 m^3 (that is riduclousy large about a million x the size of the sun)
So the radius of a sphere for a given volume is given by (3/4pi V)^1/3 which would make the radius about 7.36 x 10 ^10 m or 73600000 km (about the distance to mars)
This questions seems a non-sequitir are you sure the conc is correct?
Y'know, I got that answer, too, but apparently you should get 1.46 x 10^11m

It's not for an exam question or anything. I was trying to work out the size a spherical pill would have to be to contain one molecule of the original substance based on a 30C dilution (homeopathy stuff, if you are unfamiliar), and just compare it with something.
6. 1/(6 x 10^-37) dm^3 = volume of substance for 1 mole of substance = volume of sphere = 1.67 x 10^36
volume of sphere = 4/3 pi r^3
1.67 x 10^36 = 4/3 pi r^3
pi r^3 = 1.67 x 10^36 x 3/4 = 1.25 x 10^36
r^3 = (1.25 x 10^36)/pi = 4 x 10^35
r = 7.36 x 10^11 dm
= 7.36 x 10^10 m
= 73.6 Gm
7. This is the source, and I just wanted to try and work it through:

13:30 for the problem at hand.
8. (Original post by Artemis 97)
1/(6 x 10^-37) dm^3 = volume of substance for 1 mole of substance = volume of sphere = 1.67 x 10^36
volume of sphere = 4/3 pi r^3
1.67 x 10^36 = 4/3 pi r^3
pi r^3 = 1.67 x 10^36 x 3/4 = 1.25 x 10^36
r^3 = (1.25 x 10^36)/pi = 4 x 10^35
r = 7.36 x 10^11 dm
= 7.36 x 10^9 m
= 7.36 Gm
There are 10 dm in a meter
9. (Original post by TheDannyManCan)
Y'know, I got that answer, too, but apparently you should get 1.46 x 10^11m

It's not for an exam question or anything. I was trying to work out the size a spherical pill would have to be to contain one molecule of the original substance based on a 30C dilution (homeopathy stuff, if you are unfamiliar), and just compare it with something.
Hmm I don't see how you would get that...
Not a big fan or have that much knowledge on the alternative medicines
10. (Original post by joe3469)
There are 10 dm in a meter
oops oh yeah, sorry
11. (Original post by joe3469)
Hmm I don't see how you would get that...
Not a big fan or have that much knowledge on the alternative medicines
Me, neither! In case you didn't see, there's a link to the video above which I got this from - I think this guy is from UCL so I thought it'd be pretty reputable. I don't know =/

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