please show me how you would solve this problem!
1.000 am of dry nitrogen, placed in a container having a pinhole opening in its side, leaks from the container 3.54 times faster than does 1.000 atm of an unknown gas placed in this same apparatus. which of the following species could be the unknown gas?
answer: UF6
1.000 am of dry nitrogen, placed in a container having a pinhole opening in its side, leaks from the container 3.54 times faster than does 1.000 atm of an unknown gas placed in this same apparatus. which of the following species could be the unknown gas?
answer: UF6
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The problem is a Graham's law problem, which deals with differences in effusion rate depending on molar mass. When the question author wrote "3.54 times faster", he implied a rate, not an actual time. He was making a comparison between two rates, not two times.
You may see Graham's law written other ways, but this is the easiest to remember and to move toward a solution. (It is based on kinetic energy, KE = 1/2 m v^2)
(rate1)^2 x molar mass1 = (rate2)^2 x molar mass2
molar mass 2 = (rate1)^2 x molar mass1 / (rate2)^2
M2 = R1^2 x M1 / R2^2
M2 = 3.54^2 x 28 g/mol / 1 = 351 g/mol
Then compute the molar mass of the choices and see which one matches. UF6 has a molar mass of 350.9 g/mol.
You may see Graham's law written other ways, but this is the easiest to remember and to move toward a solution. (It is based on kinetic energy, KE = 1/2 m v^2)
(rate1)^2 x molar mass1 = (rate2)^2 x molar mass2
molar mass 2 = (rate1)^2 x molar mass1 / (rate2)^2
M2 = R1^2 x M1 / R2^2
M2 = 3.54^2 x 28 g/mol / 1 = 351 g/mol
Then compute the molar mass of the choices and see which one matches. UF6 has a molar mass of 350.9 g/mol.
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via grahams law
rate1 / rate2 = √(mw2 / mw1)
and rate = distance / t... right?
so.. assuming both gases diffuse the same distance.. "d"..
and letting "1" be nitrogen...
rate N2 = d / t
rate unknown = d / 3.54t... right? time is 3.54 times more?
mw N2 = 28.0
so..
(d / t) / (d / 3.54t) = √ ((mw unknown) / (28.0 g/mole))
(3.54) = √ ((mw unknown) / (28.0 g/mole))
(3.54)² = (mw unknown) / (28.0 g/mole)
mw unknown = (28.0 g/mole) x (3.54)² = 351 g/mole
and
molar mass UF6 = 238+6*19 = 352 g/mole
*********
questions?
You will see trick questions like this. You have to recognize that time is NOT rate... rate = d / time !
rate1 / rate2 = √(mw2 / mw1)
and rate = distance / t... right?
so.. assuming both gases diffuse the same distance.. "d"..
and letting "1" be nitrogen...
rate N2 = d / t
rate unknown = d / 3.54t... right? time is 3.54 times more?
mw N2 = 28.0
so..
(d / t) / (d / 3.54t) = √ ((mw unknown) / (28.0 g/mole))
(3.54) = √ ((mw unknown) / (28.0 g/mole))
(3.54)² = (mw unknown) / (28.0 g/mole)
mw unknown = (28.0 g/mole) x (3.54)² = 351 g/mole
and
molar mass UF6 = 238+6*19 = 352 g/mole
*********
questions?
You will see trick questions like this. You have to recognize that time is NOT rate... rate = d / time !