Question

The percentage by mole of NO2 in a mixture of NO2(g) and NO(g) having average molecular
mass 34 is
(a) 25%
(b) 20%
(c) 40%
(d) 75%
In which of the following nairs do 1 g of each have an equal number of molecules?
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Answer 1

Dear Student,

Lets assume that the mole fraction of NO2 is ‘x’ and NO is ‘y’.

Now lets assume that the container contains only these two gases. So we can say that,

x+y=1

Now we know that molecular weight of NO2 is 46 and that of NO is 30. Now,the avg molecular mass is given by

M1.x +M2.y=34

where M1 and M2 are the individual molecular masses

Now solving the two equations we find that , x= 0.25 and y= 0.75

Percentage by mole of NO2 = x * 100 = 0.25*100 = 25%

Therefore the correct option is (a) 25%

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Answer 2

Answer:25%

Explanation:let the mole fraction be x+y=1.

(mole of NO2=x and mole of NO=y)

NO2=46(wt.)

NO=30

Average mass of mixture is 34.

So,46x+30(1-x)=34

46x+30-30x=34

16x=34-30

x=4/16

=1/4

x=25%