Explanation
Moles of $$N_2 = \dfrac { 280 }{ 28 } = 10\quad moles$$
Assume, moles of $$CO_2=x$$.
So, average molecular mass of mixture $$ = \dfrac { 10\times28+x\times44 }{ 10+x } $$ $$=36$$ (given)
So, $$ x=10$$ moles
Therefore, weight of $$CO_2 = 10\times 44 = 440$$ g
Assume mole % of $$ NO_2 = x $$
Then, mole % of $$ NO = 100-x $$
So, avg. atomic mass of mixture $$ = \dfrac { x\times 46+(100-x)\times 30 }{ 100 } $$ $$= 34$$ (given)
Hence, $$x = 25 $$ %
We know that $$12\ g\ of\ Carbon=6.023\times \ 10^{23} atoms$$
So $$1\ g=\frac{6.023\ X\ 10^{23}}{12}$$
$$6\ g\ of \ carbon=6\times \frac{6.023\ X\ 10^{23}}{12}$$
or
$$=\frac{1}{2}\times 6.023\ \times 10^{23}$$
This is our new Avogadro's number according to the given conditions.
Or
$$1\ atomic\ mass\ unit=\frac{1}{N_A}=\frac{2}{6.023\ X\ 10^{23}}$$
Mass of $$1$$ mole of $$O_2=32\times \ Avogrado's\ number\times \frac{2}{6.023\ \times 10^{23}}$$
$$=32\times \frac{6.023\ \times 10^{23}}{2}\times \frac{2}{6.023\ \times 10^{23}}$$
$$=32g$$
Hence, there is no change in the mass number of $$O_2.$$
Option $$\text{C}$$ is correct.
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