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(A) $\alpha $ = 40%

(B) $\alpha $ = 10%

(C) $\alpha $ = 33%

(D) $\alpha $ = 67%

Answer
Verified

Let the initial volume of ${{N}_{2}}{{O}_{4}}$ be x , concentration of $N{{O}_{2}}$will be zero,

${{N}_{2}}{{O}_{4}}\to 2N{{O}_{2}}$

After time t at equilibrium, the concentration of ${{N}_{2}}{{O}_{4}}$ will be , $x(1-\alpha )$ and concentration of $N{{O}_{2}}$ will become, $2\alpha x$, where $\alpha $ is the de

Total initial volume = x + 0 = x

Total equilibrium volume = $x(1-\alpha )$+$2\alpha x$=$x(1+\alpha )$

According to the given in the question, the initial volume is 25% less than the final volume,

Let us assume that the initial concentration is given 100, so when it is 25% less than the final volume, it will become 75.

Therefore,

\[\begin{align} & x=75\times (1+\alpha ) \\ & \Rightarrow 100=75\times (1+\alpha ) \\

& \Rightarrow (1+\alpha )=1.33 \\

& \Rightarrow \alpha =0.33 \\

\end{align}\]

Therefore, the obtained value of degree of dissociation will be 0.33 and in terms of percentage it will be 33%.