Respuesta :
Answer:
70.77 g/mol is the molar mass of the unknown gas.
Explanation:
Effusion is defined as rate of change of volume with respect to time.
Rate of Effusion=[tex]\frac{Volume}{Time}[/tex]
Effusion rate of oxygen gas after time t = [tex]E=\frac{4.64 mL}{t}[/tex]
Molar mass of oxygen gas = M = 32 g/mol
Effusion rate of unknown gas after time t = [tex]E'=\frac{3.12 mL}{t}[/tex]
Molar mass of unknown gas = M'
The rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
[tex]\frac{E}{E'}=\sqrt{\frac{M'}{M}}[/tex]
[tex]\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}[/tex]
M' = 70.77 g/mol
70.77 g/mol is the molar mass of the unknown gas.
Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.
What does Graham's law state?
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
r ∝ 1/√M
where,
- r is the rate of effusion.
- M is the molar mass.
Under identical conditions and after the same amount of time (t), 4.64 mL of O₂ and 3.12 mL of an unknown gas X effuse. The ratio of their rates of effusion is:
[tex]\frac{rO_2}{rX} = \frac{\frac{vO_2}{t} }{\frac{vX}{t} } = \frac{vO_2}{vX} = \frac{4.64 mL}{3.12 mL} = 1.48[/tex]
We can calculate the molar mass of the unknown gas by applying Graham's law.
[tex]\frac{rO_2}{rX} = 1.48 = \sqrt{\frac{M(X)}{M(O_2)} } = \sqrt{\frac{M(X)}{32.00 g/mol} }\\\\M(X) = 70.1 g/mol[/tex]
Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.
Learn more about Graham's law here: https://brainly.com/question/22359712
