128 g/mol
Using Graham's law of effusion we have the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = rate of effusion for gas 1
r2 = rate of effusion for gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Since the atomic weight of oxygen is 15.999, the molar mass for O2 = 2 * 15.999 = 31.998
Now let's subsitute the known values into Graham's equation and solve for m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
127.992 = m2
So the molar mass of the unknown gas is 127.992 g/mol.
Rounding to 3 significant figures gives 128 g/mol
