Effusion/Diffusion Chemistry Help!!!
If 4.83mL of an unknown gas effuses through a hole in a plate in the same time it takes 9.23mL of Argon, Ar, to effuse through the same hole under the same conditions, what is the molecular mass of the unknown gas? (146amu)
Isn't this just standard use of
rate1/rate2 = sqrt(M2/M1)?
I get 146 also.
TY
To find the molecular mass of the unknown gas, we can use 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.
The equation for Graham's law is:
Rate1 / Rate2 = sqrt(Molar mass2 / Molar mass1)
In this case, we are given the rates of effusion for the unknown gas and argon:
Rate1 = 4.83 mL
Rate2 = 9.23 mL
We are also given the molar mass of argon (Ar):
Molar mass1 = molar mass of unknown gas
Molar mass2 = 39.95 g/mol (molar mass of argon)
Now, we can plug the values into the equation and solve for the molar mass of the unknown gas:
4.83 / 9.23 = sqrt(39.95 / Molar mass1)
Squaring both sides of the equation gives:
(4.83 / 9.23)^2 = 39.95 / Molar mass1
Simplifying further:
0.707^2 = 39.95 / Molar mass1
0.5 = 39.95 / Molar mass1
Next, rearrange the equation to solve for the molar mass of the unknown gas:
Molar mass1 = 39.95 / 0.5
Molar mass1 = 79.9 g/mol
Therefore, the molecular mass of the unknown gas is approximately 79.9 g/mol.