Question

At ${0^0}C$, the density of certain oxide of a gas at $2{\text{ }}bar$is same as that of dinitrogen at$5{\text{ }}bar$. What is the molecular mass of the oxide?

Answer 1

the question might seem slightly confusing, as one may think of $2{\text{ and }}5$ as the densities but they are the pressure of oxide and dinitrogen because the two are given in bar unit. Also, take into consideration that the temperature and the density of both are the same. Here the formula of the ideal gas equation in terms of density will be used.

Complete answer:
In the above question, given are the pressure of certain oxide and dinitrogen at ${0^0}C$, that is $2{\text{ }}bar$ and $5{\text{ }}bar$and we have to find the molecular mass of the oxide. It is also given that the densities of both are same. For finding out the molecular mass we will use the formula of density of the ideal gas law. The formula is given as:
$\rho {\text{ }} = {\text{ }}\dfrac{{MP}}{{RT}}$
Where $'\rho '$ is the density, ‘p’ is the pressure, ‘T’ is the temperature, ‘M’ is the molecular mass and ‘R’ is the gas constant. We can rearrange this formula as:
$P{\text{ }} = {\text{ }}\rho \dfrac{{RT}}{M}$
For the given data if M is the molar mass of the oxide, we have
$2{\text{ }} = {\text{ }}\rho \dfrac{{RT}}{M}\xrightarrow{{}}\left( 1 \right)$ Let this equation be the equation number one.
For dinitrogen we know that its molecular mass is $28.0134{\text{ }}\dfrac{g}{{mol}}$ and so for dinitrogen the equation can be written as:
$5{\text{ }} = {\text{ }}\rho \dfrac{{RT}}{{28.0134}}\xrightarrow{{}}\left( 2 \right)$ Let this equation be the equation number two.
Now from equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we have
$\dfrac{{2{\text{ }} = {\text{ }}\rho \dfrac{{RT}}{M}}}{{5{\text{ }} = {\text{ }}\rho \dfrac{{RT}}{{28.0134}}}}$
It is given that the temperature and the density of both are same and also the gas constant is same, so all the three terms can be canceled out. And so the equation becomes:
$\dfrac{5}{2} = \dfrac{M}{{28.0134}}$
$\Rightarrow \dfrac{{5 \times 28.0134}}{2} = M$
$\Rightarrow M = 70.0335{\text{ }}\dfrac{g}{{mol}}$
Hence the molecular formula of the oxide is $70.0335{\text{ }}\dfrac{g}{{mol}}$

Note:
The density of the ideal gas law is obtained from the ideal gas equation $PV = nRT$, which in terms of density is written as $PM = \rho RT$, where density is replaced by the moles and the volume is replaced by the molecular mass. And thus by rearranging this equation, the equation of density becomes: $\rho {\text{ }} = {\text{ }}\dfrac{{MP}}{{RT}}$.