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One mole of a mixture of `N_(2),NO_(2)` and `N_(2)O_(4)`, has a mean molar mass of `55.4`. On heating to a temperature at which `N_(2)O_(4)` may be dissociated `:N_(2)O_(4)rarr2NO_(2)`, the mean molar mass tends to the lower value of `39.6`. What is the mole ratio of `N_(2) : NO_(2) : N_(2)O_(4)` in the original mixture? |
Answer» Correct Answer - `0.5 : 0.1 : 0.4` Mixture `(N_(2),NO_(2),N_(2),O_(4))` has mean molar mass `=55.4` `x" "y" "z` Given: `{:(N_(2)O_(4)rarr2NO_(2)),(z" "2z):}` `therefore 55.4=(28x+46(y+2z))/(x+y+z)` `{"mean molar mass" =(wtxx"mole")/("Total mole")}` Given : `x+y+z =1` (mole) so `55.4 = 28x + 46 (y+2z)" "...(1)` `therefore 39.3 = (28x + 46(y+2z))/(x+y+2z)` `therefore 39.6(x+y+2z)=28x+46(y+2z)` From `eq (1) & x+y+z=1` or `1+z = (59.4)/(39.6)` or `z=0.4` from `eq. (1)` `55.4 = 28 x+46(y+2z)` `z=0.4` ` 55.4 = 28x + 46y +36.8` `28x+46y=18.6 " "(2)` ` because x+y+z=1` `x+y+0.4=1(beacuse z =0.4)` `x+y=0.6 " "...(3)` `eq (2) xx1....eq. (3) xx28` `{:(,28x+46y=18.6),(,28x+28y=16.8),(-, " - -"),(,bar(" 18y=1.8")):}` `y=0.1` `because x+y+z=1 x=0.5` |
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