Graham's Law of Effusion
Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically:
Rate₁ / Rate₂ = √(M₂ / M₁)
Where:
* Rate₁ and Rate₂ are the rates of effusion of gases 1 and 2, respectively
* M₁ and M₂ are the molar masses of gases 1 and 2, respectively
Applying the Law
1. Identify the known information:
* The unknown gas effuses three times slower than helium. This means the rate of helium (Rate₁) is three times faster than the rate of the unknown gas (Rate₂).
* The molar mass of helium (M₁) is 4.00 g/mol.
2. Set up the equation:
Rate₁ / Rate₂ = √(M₂ / M₁)
3 = √(M₂ / 4.00 g/mol)
3. Solve for the molar mass of the unknown gas (M₂):
* Square both sides of the equation: 9 = M₂ / 4.00 g/mol
* Multiply both sides by 4.00 g/mol: M₂ = 36.0 g/mol
Therefore, the molar mass of the gas is 36.0 g/mol.
