If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would be

Options

A. `1 : 2`
B. `2 : 5`
C. `2 : 1`
D. `1 : 2`

Answer

B

Explanation

Let mg of both the gases be enclosed in the vessel,
No. of moles of `SO_(3) = ((mg))/((80 g mol^(-1))) = (m)/(80)`
No. of moles of `O_(2) = ((mg))/((32 g mol^(-1))) = m/32`
`p_(SO_(3)) = (nRT)/(V) = (m)/(80) xx (R xx T)/(V)`
`p_(O_(2)) = (nRT)/(V) = (m)/(32) xx (R xx T)/(V)`
`(p_(SO_(3)))/(pO_(2)) = (m)/(80) xx (32)/(m) = 2/5`
The ratio of partial pressures of `SO_(3)` & `O_(2)` is `2 : 5`.