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The vapour density of a mixture consisting of NO2 and N2O4 is 38.3 at 275 K. The number of moles of NO2 in the mixture:

Option: 1

0.2


Option: 2

0.4


Option: 3

0.8


Option: 4

1.8


Answers (1)

best_answer

As we know that,

\alpha=\frac{D-d}{d}
\begin{array}{l}{D=\text { Vapour density before dissociation }} \\ {d=\text { Vapour density after dissociation }}\end{array}
          \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}
\begin{array}{l}{\text { (D) }=\frac{14 \times 2+16 \times 2}{2}=\frac{92}{2}=46} \\\\ {\text { Vapour density after dissociation }(d)=38.3} \\\\ {\therefore \alpha=\frac{46-38.3}{38.3}=0.2}\end{array}
                   \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}
initial              1                  0
at equilibrium  1-\alpha            2\alpha
\text { Number of moles of } \mathrm{NO}_{2} \text { at } \mathrm{eq},=2 \alpha=2 \times 0.2=0.4
Therefore,option(2) is correct

Posted by

Rakesh

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