J.R. S. answered 07/11/22
Ph.D. University Professor with 10+ years Tutoring Experience
Graham's law of effusion
rate gas 1 / rate gas 2 = sqrt MW gas 2 / sqrt MW gas 1
let gas 1 = 1 mol of O3
rate gas 1 = 1 mol / 9.33 min = 0.1072 mol/min
MW gas 1 = 48 g/mol
gas 2 = 1 mol of unknown
rate gas 2 = 1 mol / 5.08 min = 0.1969 mol/min
MW gas 2 = ?
0.1072 / 0.1969 min = sqrt x / sqrt 48
0.5444 = √x/48
0.2964 = x/48
x = 14.2 g/mol = molar mass of unknown gas