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Question: At the equilibrium of the reaction <img src="https://cdn.pureessence.tech/canvas_148.png?top_left_x=...

At the equilibrium of the reaction ⇌ 2NO2( g)2 \mathrm { NO } _ { 2 } ( \mathrm {~g} ) , the observed molar mass of is 77.70 g. The percentage dissociation of is

A

28.4

B

46.7

C

22.4

D

18.4

Answer

22.4

Explanation

Solution

; Molar mass of

Here, n = 2; α=92.00−77.7077.70(2−1)=0.184=18.4%\alpha = \frac { 92.00 - 77.70 } { 77.70 ( 2 - 1 ) } = 0.184 = 18.4 \%