Question

Gaseous ${{N}_{2}}{{O}_{4}}$ dissociates into gaseous $N{{O}_{2}}$ according to the reaction ${{N}_{2}}{{O}_{4}}(g)\Leftrightarrow 2N{{O}_{2}}(g)$. At 300K and 1 atm pressure, the degree of dissociation of ${{N}_{2}}{{O}_{4}}$ is 0.2. If one mole of ${{N}_{2}}{{O}_{4}}$ gas is contained in a vessel, then the density of the equilibrium mixture is:
A. $3.11g{{L}^{-1}}$
B. $6.22g{{L}^{-1}}$
C. $4.56g{{L}^{-1}}$
D. $1.56g{{L}^{-1}}$

Answer 1

You can use the formula $PV=nRT$ and derive another equation so as to introduce the density factor. After that, you can put values in the equation and get your answer.

Complete step by step answer:
As the question is talking about gases only, let us recall the famous gas equation i.e $PV=nRT$. Here, P stands for pressure, V for volume, n for number of moles, R represents the universal gas constant and the final T represents the temperature. Density is defined as the mass per unit volume of a substance. In the equation, let us write the V(volume) as mass per unit density.
So, we have
$P\dfrac{M}{d} = nRT$
Which implies
$PM = dRT$, here d is the density and we have assumed one mole of gas, so n = 1
Now let us put the respective values in the question. So we get $1\times ((94\div (1+0.2))=d\times 0.0821\times 300$, as the total number of moles is 1 + 0.2 = 1.2
So,
On solving, we get the density as 3.11 grams per litre. So this is how, by manipulating certain changes in the ideal gas equation we get the relation for density and come to our answer. So the correct answer is “A”:

Note: The equation of ideal gas $PV= nRT$ can be only used for ideal gases. It is not applicable for real gases because certain assumptions were taken in case of ideal gases, for example, that gas particles have negligible volume and that there are no intermolecular forces of attraction, which is actually not the case. So the answer that we have found in the question might not be the exact value of density, but somewhat close to it (approx.)