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?

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

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

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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?