Calculate the total vapor pressure of an aerosol mixture containing 20% CCl3F and 80% CCl2F2, given their molar masses and assuming Raoult's Law applies.

Assume a total mass of $100 \text{ g}$ for the mixture.
Mass of Propellant I ($\text{CCl}_3\text{F}$) = $20\% \text{ of } 100 \text{ g} = 20 \text{ g}$
Mass of Propellant II ($\text{CCl}_2\text{F}_2$) = $80\% \text{ of } 100 \text{ g} = 80 \text{ g}$

Using the given molar masses:
For Propellant I ($\text{CCl}_3\text{F}$, $M_1 = 137.4 \text{ g/mol}$):

$$n_1 = \frac{20 \text{ g}}{137.4 \text{ g/mol}} = 0.14556 \text{ mol}$$

For Propellant II ($\text{CCl}_2\text{F}_2$, $M_2 = 120.9 \text{ g/mol}$):

$$n_2 = \frac{80 \text{ g}}{120.9 \text{ g/mol}} = 0.66170 \text{ mol}$$

$$n_{total} = n_1 + n_2 = 0.14556 \text{ mol} + 0.66170 \text{ mol} = 0.80726 \text{ mol}$$

Mole fraction of Propellant I:

$$X_1 = \frac{n_1}{n_{total}} = \frac{0.14556 \text{ mol}}{0.80726 \text{ mol}} = 0.18031$$

Mole fraction of Propellant II:

$$X_2 = \frac{n_2}{n_{total}} = \frac{0.66170 \text{ mol}}{0.80726 \text{ mol}} = 0.81969$$

Given pure vapor pressures: $P_1^\circ = 13.4 \text{ Psia}$ and $P_2^\circ = 84.9 \text{ Psia}$.
Partial pressure of Propellant I:

$$P_1 = X_1 \times P_1^\circ = 0.18031 \times 13.4 \text{ Psia} = 2.416 \text{ Psia}$$

Partial pressure of Propellant II:

$$P_2 = X_2 \times P_2^\circ = 0.81969 \times 84.9 \text{ Psia} = 69.582 \text{ Psia}$$

$$P_{total} = P_1 + P_2 = 2.416 \text{ Psia} + 69.582 \text{ Psia} = 71.998 \text{ Psia}$$

Rounding to one decimal place:

$$P_{total} = 72.0 \text{ Psia}$$

The total vapor pressure of the aerosol mixture is $\text{72.0 Psia}$.