Question

One litre of oxygen effuses in a small hole fin 60 min and one litre of another gas xat same pressure and temp effuses through the hole in21.2 min then atmic weight of x is

Answer 1

Therefore atomic weight of gas x is 4 g/mol.

Given : One litre of oxygen effuses in a small hole fin 60 min and one litre of another gas x at same pressure and temperature effuses through the hole in 21.2 min.

To find : The atomic weight of x.

solution : according to Graham's law of effusion/diffusion, rate of effusion/diffusion is inversely proportional to square root of its molecular weight.

i.e., r ∝ 1/√M

as rate of effusion = volume of gas/time taken to effuse the given amount of gas

so, rate of effusion of oxygen gas, r₁ = 1/60 L/min

and rate of effusion of x gas, r₂ = 1/21.2 L/min

now, r₁/r₂ = √{M₂/M₁}

⇒(1/60)/(1/21.2) = √{x/32}

[ as molecular wt of oxygen gas = 32 g/mol ]

⇒21.2/60 = √(x/32)

⇒1/2.83 = √(x/32)

⇒1/(2.83)² = x/32

⇒x = 32/(2.83)² = 3.9955 ≈ 4

Therefore atomic weight of gas x is 4 g/mol.

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