If the ratio of masses of $$S{O_3}$$  and $${O_2}$$  gases confined in a vessel is 1 : 1, then the ratio of their partial pressures would be

Solution :
Let $$m$$  be the mass of $$S{O_3}$$  and $${O_2}$$  enclosed in the vessel.
Number of moles of $$S{O_3} = \frac{m}{{80}}$$
Number of moles of $${O_2} = \frac{m}{{32}}$$
Partial pressure of $$S{O_3},{P_A} = \frac{m}{{80}} \times \frac{{R \times T}}{V}$$
PartiaI pressure of $${O_2},{P_B} = \frac{m}{{32}} \times \frac{{R \times T}}{V}$$
Now, $$\frac{{{P_A}}}{{{P_B}}} = \frac{m}{{80}} \times \frac{{32}}{m} = \frac{2}{5}$$
Hence, ratio of partial pressure of $$m\,g$$  of $$S{O_3}$$  and $${O_2}$$  is 2 : 5.