Question

A mixture of Nitrogen and Oxygen has volume $3000{\text{cm}}^3$ at temperature $300\text{K}$, pressure $200\text{kPa}$ and mass $6.88\text{g}$. Then, which of the following options is/are correct?
[Take $R=8.3\text{J/mol K}$]

• A. Mass of Nitrogen in the mixture is $5.6\text{g}$.
• B. Mass of Nitrogen in the mixture is $1.28\text{g}$.
• C. Mass of Oxygen in the mixture is $1.28\text{g}$.
• D. Mass of oxygen in the mixture is $5.6\text{g}$.

Answer 1

Answer: (A), (C)

Mass of Oxygen in the mixture is $1.28\text{g}$.

Let us assume that $n_1$ and $n_2$ are the number of moles of Nitrogen and Oxygen respectively.
From Dalton's law of partial pressures, we can say that
$P=P_1+P_2$
Using ideal gas equation, we get
$P=\frac{n_{1}RT}{V}+\frac{n_{2}RT}{V}$
$=(n_1+n_2)\frac{RT}{V}$
$\Rightarrow 200\times {10}^3=(n_1+n_2)\times \frac{8.3\times 300}{3000\times {10}^{-6}}$
$\Rightarrow (n_1+n_2)=0.24.....\left(1\right)$

Given, mass of the mixture is $6.88\text{g}$.
i.e $n_1\times 28+n_2\times 32=6.88$
$\Rightarrow 14n_1+16n_2=\mathrm{3.44.......}\left(2\right)$
Now, from $\left(1\right)$,
$n_1=0.24-n_2$
Substituting the value of $n_1$ in $\left(2\right)$, we get
$\Rightarrow 14(0.24-n_2)+16n_2=3.44\text{g}$
$\Rightarrow 3.36-14n_2+16n_2=3.44\text{g}$
$\Rightarrow 2n_2=0.08$
$\Rightarrow n_2=0.04\text{mol}$
Then, from $\left(1\right)$ we get,
$n_1=0.24-0.04$
$n_1=0.2\text{mol}$
Thus,
Mass of Nitrogen $=0.2\times 28=5.6\text{g}$
Mass of oxygen $=0.04\times 32=1.28\text{g}$
Hence, options (a) and (c) are the correct answers.