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13 August, 18:33

1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?

A. CL2

B. SF6

C. Kr

D. UF6

E. Xe

Should I calcutate the molecular weight of each and usepricnipal of Grahams Law?

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  1. 13 August, 18:43
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    the gas is SF6 Sulphur hexa fluoride

    Explanation:

    According to the Graham's law of effusion or difussuion, the rate of effusion or difussion is inversely proportional to the square root of molecular mass.

    in this question it means

    mathematically: rate = k√molar mass

    rate of effusion of unknown gas/rate of effusion of Oxygen=√molar mass of unknown gas/molar mass of Oxygen

    molar mass of SF6=146.06g/mol

    =√146.06/32

    =√4.564

    =2.14.

    by trying this calculation for all the gases in the options, oxygen is 2.14 times faster than SF6.
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