Question

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

Answer 1

Answer:

b. SF₆

Explanation:

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass,

As molecular mass of O₂ is 32g/mol, the diffusion is equal to (Assuming constant of proportionality of 1):

1/√32 = 0,1768.

As this diffusion is 2,14 times faster than diffusion of the unknown gas, diffusion of the unknown gas is:

0,1768/2,14 = 0,0826

Know, this diffusion times √molec. mass of the unknown gas must be 1.

That is:

0,0826 = 1/√molec. mass

0,0826√molec. mass = 1

√molec. mass = 12,1

molecular mass of the unknown gas is: 12,1² = 146g/mol

Cl₂ has a molecular mass of 70,9 g/mol; SF₆ of 146g/mol; kr of 83,8g/mol; UF₆ of 352 g/mol and Xe of 131,3

That means the uknown gas is:

b. SF₆

I hope it helps!