Q12 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Chemistry

The vapour density of $\mathrm{N}_2 \mathrm{O}_4$ in $\mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2$ is 40 . The degree of dissociation is

  • A
    1.25
  • B
    2.50
  • C
    1.50
  • Correct answer
    D
    0.15

Solution

Given, vapour density $(d)=40$ $\mathrm{N}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{NO}_2$ Molar mass $\quad 92$ $\begin{gathered} D=\frac{M\left(\mathrm{~N}_2 \mathrm{O}_4\right)}{2}=\frac{92}{2}=46 \\ \therefore \alpha(\text { Degree of dissociation })=\frac{D-d}{(n-1) d} \end{gathered}$ $\begin{aligned} & \alpha=\frac{46-40}{(2-1) 40} \quad\left(n=\text { number of } \mathrm{NO}_2 \text { at equilibrium }\right) \\ & \alpha=\frac{6}{40}=0.15\end{aligned}$

Hence, the correct answer is option: 'D'.

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