Question

$\huge \mathfrak \red{45 \: pts}$
A mixture of ozone and oxygen containing 20% by volume of oxygen diffused through a porous plug in 172 seconds, while the same volume of pure oxygen took 164 seconds to diffuse through the same plug.
CALCULATE THE RELATIVE DENSITY OF OZONE BY USING THE EQUAL VOLUME.

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Answer 1

What is the percentage of oxygen in the mixture of the ozone and oxygen when the mixture diffused through a porous plug in 172 seconds while the same volume of pure oxygen took 164 seconds to diffuse through the same plug?

Ozone is Oxygen. It is the tri atomic form of Oxygen. So your question is meaningless. Also Ozone is unstable it can only exist for a small fraction of a second before going back to Oxygen. Ozone is O3, normal Oxygen is 02 Di atomic Oxygen.

O2 is stable , Ozone 03 is not stable. It can only exist for a few micro seconds before reverting to O2. Sorry but as it is written your question makes no sense.

108 views ·

Ozone is unstable in the lower atmosphere but, in theory, you would use Graham’s Law:

Rate∝1M√r

Replacing this with diffusion time and accounting for the fact that the volumes are the same we can say that:

tAtB=MAMB−−−√

Let mixture = A and O2= B

172164=MA32−−−√

1.01=MA32

MA=35.198

This represents the weighted mean of the Mr of the 2 gases in the mixture.

If x and y represent the fractions of the oxygen and ozone we get:

(32×x)+(48×y)=35.198(1)

x+y=1(2)

x=(1−y)

Substituting for x into (1) ⇒

32(1−y)+48y=35.198

Solving for y⇒

y = 0.20 = 20%

So the proportion of oxygen = 80 %

Answer 2

Answer:

Explanation:

What is the percentage of oxygen in the mixture of the ozone and oxygen when the mixture diffused through a porous plug in 172 seconds while the same volume of pure oxygen took 164 seconds to diffuse through the same plug?

Ozone is Oxygen. It is the tri atomic form of Oxygen. So your question is meaningless. Also Ozone is unstable it can only exist for a small fraction of a second before going back to Oxygen. Ozone is O3, normal Oxygen is 02 Di atomic Oxygen.

O2 is stable , Ozone 03 is not stable. It can only exist for a few micro seconds before reverting to O2. Sorry but as it is written your question makes no sense.

108 views ·

Ozone is unstable in the lower atmosphere but, in theory, you would use Graham’s Law:

Rate∝1M√r

Replacing this with diffusion time and accounting for the fact that the volumes are the same we can say that:

tAtB=MAMB−−−√

Let mixture = A and O2= B

172164=MA32−−−√

1.01=MA32

MA=35.198

This represents the weighted mean of the Mr of the 2 gases in the mixture.

If x and y represent the fractions of the oxygen and ozone we get:

(32×x)+(48×y)=35.198(1)

x+y=1(2)

x=(1−y)

Substituting for x into (1) ⇒

32(1−y)+48y=35.198

Solving for y⇒

y = 0.20 = 20%

So the proportion of oxygen = 80 %