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For 10 minutes each at 27^(@)C from two ...

For 10 minutes each at `27^(@)C` from two identical holes nitrogen and an unknow gas are leaked into a common vessel of 3 litre capacity The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen What is the molar mass of the unknown gas ? .

Text Solution

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Let us calculate the total number of moles in the mixture
`pV=nRT" or "n=(pV)/(RT)`
`p="4.18 bar, V = 3 L, T = 300 K,"`
`R="0.083 bar L mol"^(-1)K^(-1)`
`therefore" "n=(4.18xx3)/(0.083xx300)="0.504 mol"`
`"Moles of "N_(2)" present in the mixture = 0.4 mol"`
Moles of unknown gas diffused in the same time
`=0.504-0.4="0.104 mol"`
This means that in 10 min (same time), 0.4 mol
of `N_(2)` and 0.104 mol of unknown gas diffuse out.
As volume under similar conditions are directly proportional to their number of moles, applying Graham.s law of diffusion :
`(r_(N_(2)))/(r_(X))=sqrt((M_(X))/(M_(N_(2)))`
`((0.4)/(10))/((0.104)/(10))=sqrt((M_(X))/(28))`
`therefore" "(0.4)/(0.104)=sqrt((M_(X))/(28))`
`"or "14.8=(M_(X))/(28)`
`"or "M_(X)=14.8xx28="414.4 g mol"^(-1)`.
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