Question

2. The rate of diffusion of a gas X is root 2times that of Y. If the molecular mass of X is 16, what is the molecular mass of
gas Y​

Answer 1

Given:-

→ Rate of diffusion of gas 'X' is √2 times

that of gas 'Y'.

→ Molecular mass of 'X' = 16 u

To find:-

→ Molecular mass of gas 'Y'.

Solution:-

According to Graham's law, we know that :-

r₁/r₂ = √(M₂/M₁)

Where :-

• r₁ is the rate of diffusion of gas 'X'.

• r₂ is the rate of diffusion of gas 'Y'.

• M₂ is the molecular mass of gas 'Y'.

• M₁ is the molecular mass of gas 'X'.

According to the question :-

=> r₁ = √2r₂ ----(1)

Now by substituting values in the Graham's formula, we get :-

=> √2r₂/r₂ = √(M₂/M₁)

=> √2 = √(M₂/16)

=> √2 = √M₂/4

=> (√2)² = (√M₂/4)²

=> 2 = M₂/16

=> M₂ = 16×2

=> M₂ = 32u

Thus, molecular mass of gas 'Y' is 32u .