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

Answer

B

Explanation

`"Let mass of "SO_(3)=x," mass of "O_(2)=x`
`"Moles of "SO_(3)=(x)/(80)," Moles of "O_(2)=(x)/(32)`
`"Mole fraction of "SO_(3)=((x)/(80))/((x)/(80)+(x)/(32))`
`"Mole fraction of "O_(2)=((x)/(32))/((x)/(80)+(x)/(32))`
`"Ratio of mole fractions, "(x(SO_(3)))/(x(O_(2)))=(32)/(80)=(2)/(5)`
Since partial pressure are proportional to mole fraction
`therefore" Ratio of partial pressures "=2:5`