Question

A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 60 seconds to effuse out of the identical container under identical conditions.

What is the most likely identity of the unknown gas?
What is the most likely identity of the unknown gas?
Br2
O2
Kr
He

Answer 1

Kr

Step-by-step explanation:

Now we have to apply Graham's law of diffusion in gases

t1/t2 = √M1/M2

If t1 is the time of diffusion of Xe = 75 seconds

t2 is the time of diffusion of the unknown gas = 60 seconds

M1= molar mass of Xe=131.293

M2= molar mass of the unknown gas

Thus;

75/60 = √131.293/M2

(75/60)^2= (√131.293/M2)^2

5625/3600=131.293/M2

M2= 3600×131.293/5625

M2= 84.0

The unknown gas is krypton (Kr)

Answer 2

The unknown gas has an atomic mass of 84.0 g/mol. This is krypton (Kr)

Step-by-step explanation:

Step 1: Data given

It takes 75 seconds for Xe to effuse

Atomic mass of xenon = 131.29 g/mol

An unkknown gas (X) takes 60 seconds to effuse

Step 2: Calculate the atomic mass of the unknown gas

Time of Xe / time X = √(atomic mass Xe/ atomic gas X)

75/60 = √(131.29 / X)

1.25 = √(131.29 / X)

1.5625 = (131.29 / X)

X = 131.29 / 1.5625

X = 84.0 g/mol

The unknown gas has an atomic mass of 84.0 g/mol. This is krypton (Kr)