Multiple choice

Vapour density of the equilibrium mixture of ${ NO }{ 2 }$ and ${ N }{ 2 }{ O }{ 4 }$ is found to be 40 for the equilibrium: ${ N }{ 2 }{ O }{ 4 }\rightleftharpoons 2{ NO }{ 2 }$. Calculate the percentage of ${ NO }_{ 2 }$ in the mixture?

  1. 10%

  2. 5%

  3. 26.08%

  4. None of these
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C Correct answer
Explanation

Vapour density (VD) = M_mix / 2. M_mix = 80. Let alpha be the degree of dissociation of N2O4. M_mix = M_initial / (1 + alpha). 80 = 92 / (1 + alpha), so 1 + alpha = 92/80 = 1.15, alpha = 0.15. The mixture contains 1 mole of N2O4 and 2*alpha moles of NO2. Mole fraction of NO2 = 0.3 / 1.15 = 26.08%.

AI explanation

The molecular weight of N2O4 is 92 and NO2 is 46. Given the vapor density of the mixture is 40, its average molecular weight is calculated as 2 multiplied by 40, resulting in 80. Using the method of alligation with the molecular weights of 92 and 46 against the average of 80 gives a mole ratio of (80 minus 46) to (92 minus 80), which is 34:12 or 2.833:1. The total parts in the mixture is 3.833, so the percentage of NO2 is (1 divided by 3.833) multiplied by 100, which equals 26.08%.

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