rate of effusion of a gas is inversely proportional to the square root of the mass of its particl
v1/v2 = square root M2/M1

where:
v1 is the rate of effusion of the first gas (volume or number of moles per unit time).
v2 is the rate of effusion for the second gas.
M1 is the molar mass of gas 1
M2 is the molar mass of gas 2

in this case
Rate H2 / Rate CO2 = square root of 1/44
solve this to get the answer

Ok so I would go about this the same way I did the question that says "Find the relative rate of diffusion for the gases krypton and bromine" ?

Exactly, the exact value is not required, you only need the ratio; from that, if you know one of the exact value, the ratio can tell you the other value easily.

chemistry123

Wed, 2010-04-21 18:23

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## Re: graham's law practice problems

rate of effusion of a gas is inversely proportional to the square root of the mass of its particl

v1/v2 = square root M2/M1

where:

v1 is the rate of effusion of the first gas (volume or number of moles per unit time).

v2 is the rate of effusion for the second gas.

M1 is the molar mass of gas 1

M2 is the molar mass of gas 2

in this case

Rate H2 / Rate CO2 = square root of 1/44

solve this to get the answer

shengoc

Wed, 2010-04-21 18:24

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## Re: graham's law practice problems

graham law states that the rate of effusion(diffusion can be included) is inversely proportional to its molar mass.

ie the lighter the mass, the faster is the rate of effusion(i will just write diffusion from now on). and vice versa, which makes sense, hopefully.

so without going through all the other constants,

rate(H2)/rate(CO2) = mass(CO2) / mass(H2)

chelzluvsUM

Wed, 2010-04-21 22:01

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## Re: graham's law practice problems

Ok so I would go about this the same way I did the question that says "Find the relative rate of diffusion for the gases krypton and bromine" ?

shengoc

Thu, 2010-04-22 08:20

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## Re: graham's law practice problems

chelzluvsUM wrote:Exactly, the exact value is not required, you only need the ratio; from that, if you know one of the exact value, the ratio can tell you the other value easily.