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)) (or) (1)/(4) = (32)/(M_1)`
`M_1 = 128`
Thus the molecular mass of the unknown gas is 128.