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360 cm^3 of CH4 gas diffused through a p...

`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.

Text Solution

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Rate of diffusion of methane
`r_1=("Volume of methan e")/("Time of dif fusion of methan e")`
`=(360cm^3)/(15 min)=24cm^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 dif fusion" )`
`=(120cm^3)/(10 min)=12cm^3min^-1`
Molar mass of unknown gas `(M_2)`=?
Acording of Graham's law of diffusion,
`r_1/r_2=sqrt(M_2/M_1)`
`(24cm^3min^-1)/(12cm^3min^-1)=sqrt(M_2/(16gmol^-2))`
Squaring of both sides,
`M_2=(24 times 24 times16)/(12times12)=64`
Molar mass of the unknown gas =`64 g. mol^-1`
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