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?

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.

dsdrajlin dsdrajlin · Answer 2 · 2022-04-05T13:49:54+02:00

1.000 atm oxygen gas effuses 2.14 times faster than 1.000 atm SF₆.

What does Graham's law state?

Graham's law states that the rates of diffusion and effusion of a gas are inversely proportional to the square root of the molar mass of the gas.

1.000 atm oxygen gas effuses 2.14 times faster than the unknown gas X. We can calculate the molar mass of gas X using Graham's law.

vO₂/vX = √{M(X)/M(O₂)]

2.14 = √{M(X)/32.00 g/mol]

M(X) = 146.5 g/mol

We have the following gases with their respective molar masses.

A. Cl₂ (70.90 g/mol)
B. SF₆ (146.05 g/mol)
C. Kr (83.80 g/mol)
D. UF₆ (352.02 g/mol)
E. Xe (131.29 g/mol)

Due to its molar mass, SF₆ is most likely the unknown gas X.

1.000 atm oxygen gas effuses 2.14 times faster than 1.000 atm SF₆.

Learn more about effusion here: https://brainly.com/question/2097955