Question

He and a gas X having the same volume, diffuse through a porous partition is 15 and 30 seconds respectively. The gas X can be​

Answer 1

Explanation:

√m2/√m1=t2/t1

√x/√4=30/15

√x/√4=2

x/4=4

x=16(oxygen gas)

Answer 2

Answer:

The 'X' gas is oxygen having a molar mass of $16g$.

Explanation:

Given,

The diffusion time of $He$ gas, T₁ = $15 seconds$

The diffusion time of 'X' gas, T₂ = $30 seconds$

The volume of both gases is the same.

The gas 'X' =?

As we know,

• According to Graham’s Law of diffusion, the rate of diffusion is inversely proportional to the molecular mass of the gas.
• Rate of diffusion $\alpha \frac{1}{\sqrt{Molar mass} }$

Therefore,

• $\frac{R_{1} }{R_{2} } = \sqrt{\frac{M_{2} }{M_{1} } }$    -------equation (1)

Here,

• R₁ = Rate of diffusion of gas 1
• R₂ = Rate of diffusion of gas 2
• M₂ = Molar mass of gas 1
• M₁ = Molar mass of gas 2

As we know,

• Rate = $\frac{Volume}{Time}$

As given, volume is constant, therefore:

• $R\alpha \frac{1}{T}$
• $\frac{R_{1} }{R_{2} } = \frac{T_{2} }{T_{1} }$

After putting this in equation (1), we get:

• $\frac{T_{2} }{T_{1} } = \sqrt{\frac{M_{2} }{M_{1} } }$   -------equation (2)

Here,

• T₂ = Diffusion time of gas 1, i.e., Helium
• T₁ =  Diffusion time of gas 2, i.e., 'X' gas
• M₂ = Molar mass of gas Helium = $4g/mol$
• M₁ = Molar mass of gas 'X'

After putting the given values in equation (2), we get:

• $\frac{30 }{15 } = \sqrt{\frac{M_{X} }{4 } }$
• $\frac{M_{X} }{4} = (2)^{2}$
• $M_{X}= 16g$

The molar mass of gas 'X' = $16g$

As we know, oxygen gas has a molar mass of $16g$.

Therefore, the 'X' gas is oxygen.