Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant pressure conditions. It required 67 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

Answer 1

Sec het molar mass of un known gas (M) = re almal 6 12 gas diffuses in 67 sec. rate of diffusion of unknown des = = Oools molar mass of o2 = 32 gimol Al gas diffuses in 31 sec. ate of diffusion of o2+ i = 0-032 sec. rate of diffusione a mostanmess . a rate of diffusion of oz Munknowhaal sion of unknongal Moz sate ne = 00032 0.015 a ze=149.5 32 (molar mass of unknown gas = = luges g) mol CScanned with CamScanner