Question

Vapour density of a mixture of NO2 and N2O4 is 34.5 then percentage abundance of NO2 in mixture is
1)50%
2)25%
3)40%
4)60%

Answer 1

Answer:

Explanation:

=> Here, mixture of NO2 and N2O4 is given. So,

                  N2O4    →    2NO2

initial moles;      1            0

at equilibrium;   1−x         2x

total moles at equilibrium = 1−x +2x = 1+x

Molecular weight of equilibrium mixture = 2 * 34.5 (V.D.) = 69

Molecular weight at initial = 92

=> Thus, the value of x can be calculated as:

initial moles / moles at equilibrium = M.wt. of equilibrium mixture / Initial M.wt.

1 / 1+x = 69 / 92

1+x = 92/69

1+x = 1.33

x = 1.33 - 1

x = 0.33

Total moles at equilibrium = 1 + x = 1 + 0.33 = 1.33

Moles of NO2 at equilibrium = 2x = 2*0.33 = 0.66

=> Percentage abundance of NO2 = moles of NO2 at equilibrium / total moles at equilibrium

= 0.66/1.33 * 100

= 0.4962 * 100

= 49.62%

≈ 50%

Hence,  percentage abundance of NO2 in mixture is approx 50%.

Answer 2

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=> Here, mixture of NO2 and N2O4 is given. So,

                  N2O4    →    2NO2

initial moles;      1            0

at equilibrium;   1−x         2x

total moles at equilibrium = 1−x +2x = 1+x

Molecular weight of equilibrium mixture = 2 * 34.5 (V.D.) = 69

Molecular weight at initial = 92

=> Thus, the value of x can be calculated as:

initial moles / moles at equilibrium = M.wt. of equilibrium mixture / Initial M.wt.

1 / 1+x = 69 / 92

1+x = 92/69

1+x = 1.33

x = 1.33 - 1

x = 0.33

Total moles at equilibrium = 1 + x = 1 + 0.33 = 1.33

Moles of NO2 at equilibrium = 2x = 2*0.33 = 0.66

=> Percentage abundance of NO2 = moles of NO2 at equilibrium / total moles at equilibrium

= 0.66/1.33 * 100

= 0.4962 * 100

= 49.62%

≈ 50%

Hence,  percentage abundance of NO2 in mixture is approx 50%.

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