An unknown gas effuses at a speed one quarter of that of helium. what is the molar mass of the unknown gas? it is either sulfur di oxide or sulfur tri oxide. which gas is it.

sulfur dioxide.

Explanation:

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).

∵ 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.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.