Question

Calculate the density of Freon \(12,{\rm{C}}{{\rm{F}}_2}{\rm{C}}{{\rm{l}}_2}\), at \({30.0^\circ }{\rm{C}}\) and \(0.954\;{\rm{atm}}\).

Short answer

Density of Freon \(12,{\rm{C}}{{\rm{F}}_2}{\rm{C}}{{\rm{l}}_2}\), at \({30.0^\circ }{\rm{C}}\) and \(0.954\;{\rm{atm}}\) is \(4.63\frac{{{\rm{ gram }}}}{L}\).

Step-by-step solution

Definition of density

The density of a material is a measurement of how densely it is packed together. The mass per unit volume is how it's defined.

Define density and \(\mu \)

Let us solve the given problem.

We have,

\(PV = nRT\)

Where,

\(P = \)Pressure of the gas.

\(V = \)Volume of the gas.

\(n\)= number of moles of the gas.

\(T = \)Temperature of the gas.

\(\begin{aligned}{}R &= {\rm{ Ideal gas constant}}\\ &= 0.08206\;{\rm{L atm mo}}{{\rm{l}}^{ - 1}}{K^{ - 1}}\end{aligned}\)

If, amount of the gas\((n)\)is in moles, temperature\((T)\)is in Kelvin\((K)\), pressure\((P)\)is in atm.

\(\begin{aligned}{}\frac{P}{{RT}} &= \frac{n}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{{n \cdot (\mu )}}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{m}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \rho \end{aligned}\)

Density\((\rho ) = \frac{{P \cdot (\mu )}}{{RT}}\)

Where\(\mu = \)molar mass.

Converting temperature

Consider the given information.

Pressure\((P) = 0.954{\rm{kPa}}\).

Temperature\((T) = {30^\circ }C\).

Molar mass\((\mu )\)of\({\bf{C}}{{\bf{F}}_2}{\bf{C}}{{\bf{l}}_2}\)

Atomic weight of Carbon\(({\rm{C}}) = 12\)grams.

Atomic weight of Florine\((F) = 19\)gram.

Atomic weight of Chlorine\(({\rm{Cl}}) = 35.45\)gram.

So, molecular weight of\({\rm{C}}{{\rm{F}}_2}{\rm{C}}{{\rm{l}}_2} = 12 + (2) \cdot (19) + (2) \cdot (35.45)\)grams.

\(\begin{aligned}{} = 12 + 38 + 70.9\\ = 120.9\,grams.\end{aligned}\)

Molar mass of\({\rm{C}}{{\rm{F}}_2}{\rm{C}}{{\rm{l}}_2} = 120.9\frac{{{\rm{ gram }}}}{{{\rm{ mol }}}}\)

Converting the temperatures in degree Celsius to kelvins.

We have,\({0^\circ }C = (0) + 273.15K\)

Therefore,

\(\begin{aligned}{}{30^\circ }C = (30) + 273.15K\\ = 303.15K\,.\end{aligned}\)

Find density

Calculate density in order to get the final answer.

\(\begin{aligned}{}{\mathop{\rm Density}\nolimits} (\rho )& = \frac{{P \cdot (\mu )}}{{RT}}\\ &= \frac{{(0.954) \cdot (120.9)}}{{(0.08206) \cdot (303.15)}}\\ &= \frac{{115.338}}{{24.876}}\end{aligned}\)

&=\(4.63\frac{{{\rm{ gram }}}}{L}\).

Therefore, the calculated density is\(4.63\frac{{{\rm{ gram }}}}{L}\).