Turn on thread page Beta
    • Thread Starter
    Offline

    6
    ReputationRep:
    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!
    Offline

    0
    ReputationRep:
    I'm not entirely sure, but wouldn't it depend on the density of the substance? as density=mass/volume?
    Offline

    0
    ReputationRep:
    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?
    • Thread Starter
    Offline

    6
    ReputationRep:
    I'm not sure. It's being diluted in water, density of which is 1kg per litre, I think.
    • Thread Starter
    Offline

    6
    ReputationRep:
    (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.
    Offline

    1
    ReputationRep:
    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
    • Thread Starter
    Offline

    6
    ReputationRep:
    This is the source, and I just wanted to try and work it through:
    http://www.youtube.com/watch?v=YMvIrYEtmOM

    13:30 for the problem at hand.
    Offline

    0
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    (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
    Offline

    1
    ReputationRep:
    (Original post by joe3469)
    There are 10 dm in a meter
    oops oh yeah, sorry
    • Thread Starter
    Offline

    6
    ReputationRep:
    (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 =/
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 2, 2011
Poll
Cats or dogs?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.