Question

A container contains mixture of o2 and h2. If mole fraction of h2 is 1/3 then the ratio of volume of h2 to o2 diffuse out initially from the container is a. 1:2 b. 2:1 c. 1:1 d.4:1

Answer 1

Given:

The mole fraction of H2, X(H2) = 1/3

To Find:

The ratio of the volume of H2 to O2 that diffuses out initially from the container.

Calculation:

- The mole fraction of O2 = 1 - Mole fraction of H2

⇒ X(O2) = 1 - 1/3 = 2/3

- We know that equal no of moles of gases occupy equal volumes.

- Let the volume of the container be V.

⇒ The volume of H2, V1 = V/3

& The volume of O2, V2 = 2V/3

- The rate of diffusion = r1/r2

⇒ r1/r2 = (V1/V2) × √(M2/M1)

⇒ r1/r2 = (V/3)/(2V/3)) × √(32/2)

⇒ r1/r2 = 1/2 × 4

⇒ r1/r2 = 2:1

⇒ Ratio = 2 : 1

- Hence, initially the diffusing hydrogen to oxygen volume ratio is 2:1.

- So, the correct answer is option (b) 2 : 1.

Answer 2

A container contains mixture of o2 and h2. If mole fraction of h2 is 1/3 then the ratio of volume of h2 to o2 diffuse out initially from the container is a. 1:2 b. 2:1 c. 1:1 d.4:1

Explanation:

rate of diffusion =

r 2/r 1= V 1/V2 × M 1/M 2

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= 2/1 × 32/2

= 8/1

The answer is 1:4

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