Question

At constant temperature, He effuses 6.04 times faster than a gas having an unknown molar mass. Determine the molar mass of the unknown gas.

Answer 1

According to Graham's law of effusion, we can solve for the molar mass of the unknown gas by comparing its rate of effusion to that of helium. The molar mass of the unknown gas is 146.4 g/mol.

Step-by-step explanation:

Graham's law of effusion relates the rate of effusion of a gas to its molar mass. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In this case, helium (He) effuses 6.04 times faster than the unknown gas at constant temperature. Therefore, we can set up the following equation:

(Rate of effusion of He) / (Rate of effusion of unknown gas) = sqrt((Molar mass of unknown gas) / (Molar mass of He)), where Molar mass of He is 4.0026 g/mol.

Substituting the given values, we have:

(1) / (6.04) = sqrt((Molar mass of unknown gas) / (4.0026)),

Simplifying further, we can solve for the molar mass of the unknown gas:

(Molar mass of unknown gas) = 4.0026 x (6.04)^2 = 146.4 g/mol.

Therefore, the molar mass of the unknown gas is 146.4 g/mol.

Answer 2

Step-by-step explanation:

Rate of effusion =

`Rate of effusion = (k)/(√(M) )`

k is a constant and M is molecular mass .

Molecular mass of He = 4 .

rate of effusion of He = r₁

Molecular mass of unknown compound = M

rate of effusion of unknown gas = r₂

r₁ = k / √4

r₂ = k / √M

r₁ / r₂ = √ ( M/4 )

6.04 = √ ( M/4 )

36.48 = M/4

M = 146 approx .

molecular mass of unknown gas = 146 .