Question

What is the density of Freon-$11$ ($CFC{l_3}$) at $166$ degrees Celsius and $5.92atm$ ?

Answer 1

Hint: The ideal gas law, also known as the general gas equation, is a hypothetical ideal gas's equation of state. While it has many drawbacks, it is a reasonable approximation of the action of certain gases under several conditions.

Complete answer:
Trichlorofluoromethane is a chlorofluorocarbon also known as freon-$11$ and its chemical formula is $CFC{l_3}$. It is a colourless, sweetish-smelling liquid that melts at room temperature.
According to the question the density of $CFC{l_3}$has to be found out. To find this, the ideal gas law can be used.
The Ideal gas law states that:
$pV = nRT$
Here,
$p$is the absolute pressure of a gas,
$V$is the volume it occupies,
$n$is the number of atoms and molecules in the gas
And $T$is its absolute temperature.
We know that $n = \dfrac{m}{M}$, $M$is the molar mass and $m$is the mass of the substance measured in the grams.
Substituting the above equation in the gas law, we get:
$pV = \dfrac{m}{M}RT$
$ \Rightarrow pM = \dfrac{m}{V}RT$
We know that the density of the compound $\rho = \dfrac{m}{V}$. Thus, by substituting this relation in the above equation becomes:
$ \Rightarrow pM = \rho RT$
Therefore,
$ \Rightarrow \rho = \dfrac{{pM}}{{RT}}$
Substituting the values from the problem which are,
$p = 5.92atm$
$M = 137.37g$/$mol$
$R = 0.08206Latm{K^{ - 1}}mo{l^{ - 1}}$
$T = {166^ \circ }C = 439.15K$
Thus,
$\rho = \dfrac{{5.92 \times 137.37}}{{0.082 \times 439.15}}g$/$L$$ = 22.6g$/$L$
Thus, the density of $CFC{l_3}$at $166$ degrees Celsius and $5.92atm$is $22.6g$/$L$.

Note:
The concept "ideal gas" describes a hypothetical gas made up of molecules that adhere to a set of rules: Ideal gas molecules are neither attracted nor repellent to one another. The only contact with ideal gas molecules will be an elastic collision as they collided with each other or with the container's walls.