Question

6.

A given volume of nitrogen gas required 68.3 seconds to diffuse through a tiny hole
in chamber. Under the same conditions, another unknown gas requires 85.6 seconds for
the same volume to diffuse. What is the molecular mass of this gas?​

Answer 1

Molar mass of the unknown gas is 8.96g/mol

Step-by-step explanation:

Hello,

According to Graham's law of diffusion, its states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass

R = k (1/√(M))

R = rate of diffusion

M = molecular mass

R₁ / R₂ = √(M₂ / M₁)

R₁ = nitrogen = 68.3s

R₂ = unknown gas = 85.6s

M₂ = molecular mass of the unknown gas

M₁ = molecular mass of nitrogen = 14g/mol

68.3 / 85.6 = (√M₂ / 14)

0.8 = (√M₂ / 14)

Take square of both sides

0.8² = M₂ / 14

M₂ = 14 × 0.64

M₂ = 8.96g/mol

The molar mass of the unknown gas is 8.96g/mol