Question

Determine the density of chlorine gas at ${22.0^0}C$ at $1$ atm?

Answer 1

According to the ideal gas law, the product of volume and temperature is equal to the number of moles, universal gas constant, and temperature. Whereas the density is the ratio of mass and volume. From these two definitions, the ratio of pressure to the product of universal gas constant and temperature gives the volume.
Formula used:
$PV = nRT$
P is pressure
V is volume
n is number of moles
R is universal gas constant
T is temperature
$d = \dfrac{m}{V}$
d is density
m is mass
V is volume

Complete answer:
Let us consider the equations of ideal gas law, and the density.
$PV = nRT$ whereas n is the number of moles. It is the ratio of mass to molar mass.
Thus, it can be written as $n = \dfrac{m}{M}$ where m is the mass of the chlorine gas and M is the molar mass gas.
From the definition of moles, and ideal gas law, $PV = \dfrac{m}{M}RT$
Now, the density is nothing but the mass to volume i.e.., $d = \dfrac{m}{V}$
Thus, from the above two equations it can be written as $\dfrac{P}{{RT}} \times M = \dfrac{m}{V} = d$
Given that the temperature is ${22.0^0}C$ and pressure is $1$ atm, the universal gas constant is $0.0821$ and the molar mass is $71amu$.
By substituting these values, the density of chlorine gas is $2.93g/L$ .
Thus, the density of chlorine gas is $2.93g/L$

Note:
While calculating any term or quantity related to ideal gas law, the units must be considered. As the pressure must be in atm, and the volume must be in Litres, the temperature must be in kelvins, and number of moles in moles only.