 # 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 6 years ago
R

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

________________

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

________________

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!