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Question

If vapour density of equilibrium $$ N_{2}O_{4}(g)\leftrightharpoons 2NO_{2}(g),$$ is found to be 31 then the degree of dissociation of $$ N_2O_4 $$ will be:


A
0.48
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B
0.25
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C
0.65
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D
0.35
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Solution

The correct option is C 0.48
$${ N }_{ 2 }{ O }_{ 4 }\rightleftharpoons { NO }_{ 2 }$$
    $$1$$              $$-$$
$$1-x$$          $$2x$$
$$\dfrac { { n }_{ initial } }{ { n }_{ final } } =\dfrac { { M }_{ final } }{ { M }_{ initial } } \quad \quad { M }_{ final }=2\times VO=32$$
$$\dfrac { 1 }{ 1+x } =\dfrac { 62 }{ 92 } \Rightarrow x=0.48$$

Chemistry

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