Vapour density of a mixture of NO2 and N2O4 is 34.5 then percentage ab

1.	Vapour density of a mixture of NO2 and N2O4 is 34.5 then percentage abundance of NO2 in mixture is1)50%2)25%3)40%4)60%please give step by step solution
Answer» 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 = 20.33 = 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%.

Vapour density of a mixture of NO2 and N2O4 is 34.5 then percentage abundance of NO2 in mixture is1)50%2)25%3)40%4)60%please give step by step solution

Answer»

Answer:

Explanation:

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

N2O4 → 2NO2

initial moles; 1 0

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%.