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At 340 K and 1 atm pressure, N(2)O(4) is...

At `340 K` and `1` atm pressure, `N_(2)O_(4)` is `66%` into `NO_(2)`. What volume of `10 g N_(2)O_(4)` ocuupy under these conditions?

Text Solution

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`{:(,N_(2)O_(4),hArr,2NO_(2)),("Mole before eqm.",1,,0), ("Mole after eqm.",(1-alpha),,2alpha):}`
Where `alpha` is degree of dissociation.
`:. ` Total mole at equilibrium `=1+alpha=1+0.66` (`:' alpha=0.66`)
`=1.66`
`1 "mole"` of `N_(2)O_(4)` is taken, mole at equilibrium `=1.66`
`(10)/(92) "mole"` of `N_(2)O_(4)` is taken, mole at equilibrium
`=(1.66xx10)/(92)=0.18`
Now, `PV=(w)/(m)RT`
`implies V=(wRT)/(P)=(0.18xx0.082xx340)/(1)`
`=5.01~~5.0 litre`
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