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At 60^{\circ} \mathrm{C} the density of \mathrm{N}_2 \mathrm{O}_4 gas was found to be 30.2. The percentage of \mathrm{NO}_2 molecules -

Option: 1

68.6 \%


Option: 2

70 \%


Option: 3

60 \%


Option: 4

98.6 \%


Answers (1)

best_answer

\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2
1-\alpha           2\alpha = 1+\alpha

If undissociated density \mathrm{\rho_0=\frac{M}{V}}
After dissociation density  \mathrm{\rho=\frac{M}{(1+\alpha) V}}

\mathrm{\therefore \frac{\rho}{\rho_0}=\frac{1}{1+\alpha} \Rightarrow \rho_0 =\rho(1+\alpha)}
\mathrm{\therefore \alpha =\frac{\rho_0-\rho}{\rho}=\frac{46-30.2}{30.2}=0.523}
\mathrm{\therefore\, Percentage\, of \, \mathrm{NO}_2\, molecules}

\mathrm{=\frac{2 \alpha \times 100}{1+a}=\frac{2 \times 0.523 \times 100}{1+0.523}=68.6 \%}

Option (a) correct.

Posted by

Irshad Anwar

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