1 mole of a mixture of N2,NO2,N2O4 has a mean molar mass of 55.4on heating to a temperature at which all the N2O4 may be pressured to have dissociated as N2O4..... 2NO2,the mean molar mass becomes 39.6. what is the mole ratio of N2,NO2,N2O4 in original mixture

Let, No. of moles of N2 = x

No. of moles of NO2 = y

So, no. of moles of N2O4 = Z

So, we get, x + y + z = 1----eq.1

Now. as ,mean average of the molar mass of all these molecules = 55.4

So, we can create an equation as: Molar mass of N2 = 28 , molar mass of  NO2= 46 and Molar mass of  N2O4 = 92

So, we get, [28x + 46 y + 92z ]/ [x +y +z] = 55.6

or, x+y+z=1

So,  28x + 46 y + 92z = 55.6.... eq.2 

Further, since on heating all the N2O4 is converted to 2NO2 so moles of NO2 formed is doubled of moles of N2O4 so 

moles of NO2 formed= 2z

hence, in new condition moles of NO2 = (y +2z)

Now, mixture has x moles of N2 and (y+2z) moles of NO2 (i.e. , y moles of NO2 =46y and 2z moles of NO2 = 92z)so, the mean molar mass decreases as :

[28x+ 46y + 92 z ]/ (x + y +2z) = 39.6

Or, putting the value from eq.2,

x + y + 2z = 55.4/39.6 = 1.39...eq.3

Now we subtract  eq.1 from 3

 x + y  +2z= 1.39

-x  -y  -z= -1  

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z = 0.39 = 0.4 

So, x+y = 0.6 

Putting the value of z in eq.2 

28x + 46 y = 18.8

-x     - y   = - 0.6 ] x 28

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18y = 2

y = 2/18 =0.1

hence, x = 0.6 - 0.1 = 0.5

So, ratio becomes , x : y : z = 0.5 : 0.1 : 0.4 

I hope, it is clear now!