Question

Let the number of moles of ( mathrm{N}_{2} ) be ( mathrm{X} ), the number of moles of ( mathrm{NO}_{2} ) be ( mathrm{y} ). and number of moles of ( mathrm{N}_{2} mathrm{O}_{4} ) be ( mathrm{Z} ).
Total moles ( =1 )
So ( x+y+z=1 ) Equation 1
Molar mass of ( mathrm{N}_{2}=28 )
Molar mass of ( mathrm{NO}_{2}=46 )
Molar mass of ( mathrm{N}_{2} mathrm{O}_{4}=92 )
Mean average of the molar mass ( =frac{28 x+46 y+92 z}{x+y+z}=55.6 )
Using the relation ( x+y+z=1 )
( 28 x+46 y+92 z=55.6 ) Equation 2
moles of ( mathrm{NO}_{2} ) formed from ( mathrm{N}_{2} mathrm{O}_{4}=2 mathrm{z} )
Hence after reaction, ( mathrm{NO}_{2}=mathrm{y}+2 mathrm{z} )
Now mean average of the molar mass after reaction ( =frac{28 x+46 y+92 z}{x+y+2 z}=39.6 ) Equation 3
Solving the three equations
We get
( x=0.5 )
( y=0.1 )
( z=0.4 )
Thus the ratio becomes ( x: v: z=0.5: 0.1: 0.4 )

# 1. 50, 000 Ms. mole of mixture of Ng. NO, and N, O, has a mean molar mass of 55.4 x on which all the N, O, may be dissociated into NO, the mean molar mass tends What is the mole ratio of NNO, and N, O, in the original mixture? e molar mass of the protein ins (MA2 (D) 12.00.000 On heating to a temperature. ass tends to a lower value of 39.6 g. (A) 5:1:4 A protein, isolated from a bovine © 1:1:1 (D) 1:5 : 4

Solution