Question

Calculate density of a gaseous mixture which consist of \[3.01 \times {10^{24}}\] molecules of \[{N_2}\] and \[32g\] of \[{O_2}\] gas at \[3{\text{ }}atm\] pressure and \[860K\] temperature (Given: \[R = 1/12{\text{ }}atm{\text{ }}L/mole.K\])
A.\[0.6g/L\]
B.\[1.2g/L\]
C.\[0.3g/L\]
D.\[12g/L\]

Answer 1

Hint: Density is a measure of mass per volume. The average density of an object equals its total mass divided by its total volume. An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas.

Complete answer:
For nitrogen:
Number of nitrogen molecules \[ = 3.01 \times {10^{24}}\]
According to the mole concept, there are \[6.022 \times {10^{23}}\] number of atoms or molecules present in \[1\] mole of a given element or compound. Therefore, the number of moles of nitrogen molecules are \[ = 3.01 \times {10^{24}} \times \dfrac{1}{{6.022 \times {{10}^{23}}}}\]
\[ = 5\] mole of nitrogen molecule.
We know that, to calculate the mass of any compound for a given number of moles, the following relation is used:
Number of moles \[ = \dfrac{{given{\text{ }}mass}}{{molar\;mass}}\]
The molar mass of nitrogen is \[28g/mol\] and the number of moles present here is \[5\] moles.
Hence, given mass \[ = \] number of moles \[ \times \] molar mass
Given mass \[ = 28g/mol \times 5mol = 140g\]
For oxygen:
Given mass of oxygen \[ = 32g\]
Molar mass of oxygen \[ = 32g/mol\]
Therefore, Number of moles \[ = \dfrac{{given{\text{ }}mass}}{{molar\;mass}}\]
Number of moles \[ = \dfrac{{32g}}{{32g/mol}} = 1mol\]
According to ideal gas equation:
\[PV = nRT\]
\[P \to \] pressure of gas \[ = 3atm\]
\[V \to \] volume of gas (to be calculated)
\[n \to \] number of moles of mixture \[ = 5 + 1 = 6\]
\[R \to \] gas constant \[ = \dfrac{1}{{12}}L{\text{ }}atm{\text{ }}mol{e^{ - 1}}{K^{ - 1}}\]
\[T \to \] temperature of gas \[ = 860K\]
Hence, \[V = \dfrac{{nRT}}{P}\]
\[V = \dfrac{1}{{3atm}} \times 6mol \times \dfrac{1}{{12}}L\;atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}} \times 860K\]
\[V = 143.33L\]
Now, density of the gas can be calculated as follows:
Density \[ = \dfrac{{{\text{mass}}\;{\text{of}}\;{\text{substance}}}}{{{\text{volume}}\;{\text{of}}\;{\text{substance}}}}\]
Here, mass of mixture \[ = 140 + 32 = 172g\]
And, volume of mixture \[ = 143.33L\]
Hence, density \[ = \dfrac{{172g}}{{143.33L}}\]
Density \[1.2g/L\]
Hence, the correct option is B.

Note:
An object made from a comparatively dense material (such as iron) will have less volume than an object of equal mass made from some less dense substance (such as water). The ideal gas law is a good approximation of the behaviour of many gases under many conditions, although it has several limitations like it does not tell us whether a gas heats or cools during compression or expansion.