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The vapour density of N2O4 at a certain temperature is 30. Calculate the percentage dissociation of N2O4 at this temperature.
 

Option: 1

85.3%


Option: 2

63.6%


Option: 3

20%


Option: 4

53.3%


Answers (1)

best_answer

\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})
\begin{array}{l}{\text { Molecular mass of } \mathrm{N}_{2} \mathrm{O}_{4}=(28+64)=92} \\ {\text { Vapour density, } \mathrm{D}=\frac{92}{2}=46} \\ {\text { Let the degree of dissociation be } x .} \\ {\text { Given, }\mathrm {d}=30}\end{array}
\begin{array}{l}{\text{{Applying the relationship,} }}\\\\ {x=\frac{D-d}{d}=\frac{(46-30)}{30}=\frac{16}{30}=0.533} \\\\ {\text { Degree of dissociation }=53.3 \%}\end{array}

Therefore,option(4) is correct

Posted by

Divya Prakash Singh

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