If a gas diffuses at the rate of one-half as fast as `O_(2)`, find the molecular mass of the gas.

Explanation

Applying Graham's law of Diffusion,
`(r_(1))/(r_(2))= sqrt((M_(2))/(M_(1))), (1//2)/(1)= sqrt((32)/(M_(1))`
Squaring both sides of the equation,
`((1)/(2))^(2)=(32)/(M_(1)), (1)/(4)=(32)/(M_(1))`
`:. M_(1)=128`
Thus the molecular mass of the unknown gas is 128.