Answer:
sulfur dioxide.
Explanation:
- Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.
Rate of effusion ∝ 1/√molar mass.
(Rate of effusion of He) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of He).
- An unknown gas effuses at a speed one quarter of that of helium.
∵ Rate of effusion of unknown gas = 1/4 (Rate of effusion of He)
∴ (Rate of effusion of He) / (Rate of effusion of unknown gas) = 4.
Molar mass of He = 4.0 g/mol.
∵ (Rate of effusion of He) / (Rate of effusion of unknown gas) = (√molar mass of unknown gas) / (√molar mass of He).
∴ 4.0 = (√molar mass of unknown gas) / √4 = (√molar mass of unknown gas) / 2.
∴ (√molar mass of unknown gas) = 8.0.
By squaring the both sides:
∴ molar mass of unknown gas = 64.0 g/mol.
The molar mass of sulfur dioxide = 64.0 g/mol and the molar mass of sulfur trioxide = 80.0 g/mol.
So, the unknown gas is sulfur dioxide.
