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Q)

An oxide of nitrogen at NTP has a density equal to another oxide of Nitrogen at $12^{\circ}$. If we consider pressure conditions to be the same, what are the two oxides involved?

(A) $NO, NO_2$ (B) $N_2O, NO_2$ (C) $NO, N_2O$ (D) $NO_2, NO$

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A)
For any ideal gas, density $\rho = \large\frac{PM}{RT}$
So, for densities to be equal under similar conditions of pressure, $\large\frac{M}{T}$ needs to be the same.
For $N_2O$ and $NO_2, \; \large\frac{44}{273} $$= \large\frac{46}{273+12}$
Hence this is the correct answer. (We can verify the values of $\large\frac{M}{T}$ for the other combinations and we'll see that they are not the same).
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