1.

At `77^(@)C` and one atmospheric pressure , `N_(2)O_(4)` is 70% dissociated into `NO_(2)` What will be the volume occupied by the mixture under these conditions if we start with 10 g of `N_(2)O_(4)` ?

Answer» Molar mass of ` N_(2)O_(4) = 28 + 64 = 92 g "mol"^(-1)`
` {:(,N_(2)O_(4),hArr,2NO_(2)),("Intial moles",10/12,,0),("After dissociation",10/92-70/100xx10/92,,2xx0*076),(,=0*109 - 0*076=0*033,,= 0*152):}`
` :. " Total moles after dissociation " = 0*033 + 0* 152 = 0*185, T = 77^(@)C = 77 + 273 K= 350 K`
` Pv = nRT or V= (nRT)/P = ( 0*185 "mole" xx 0* 0821 L "atm " K^(-1) "mol"^(-1) xx350 K )/(1atm) = 5* 32 L.`


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