`360 cm^3` of `CH_4` gas diffused through a porous membrane in 15 minutes. Under similar conditions, `120cm^3` of another gas diffused in 10 minutes. Find the molar mass of the gas.

Explanation

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or vapour pressure or molecular weight .
Methane
Rate of diffusion of methane
`r_(1)=("Volume of methane")/("Time of diffusion of methane")=(360cm^(3))/(15"min".)=24"cm"^(3)"min"^(-1)`
Molar mass of `CH_(4)(M_(1))=16"gmol"^(-1)`
Unknown Gas
Rate of diffusion of unknown gas
`r_(2)=("Volume of unknown gas")/("Time of diffusion")=(120cm^(3))/(10"min".)=12"cm"^(3)"min"^(-1)`
Molar mass of unknown gas `(M_(2))` = ?
According to Graham.s law of diffusion,
`(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`
`(24"cm"^(3)"min"^(-1))/(12"cm"^(3)"min"^(-1))=sqrt((M_(2))/(16"gmol"^(-1)))`
Squaring on both sides,
`M_(2)=(24xx24xx16)/(12xx12)=64`
Molar mass of the unknown gas = `64 "g. mol"^(-1)`