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

Answer

C

Explanation

According to Dalton's law of partial pressures,
`p_(SO_(3)) = (n_(SO_(3)))/(n_(T)) . p_("total")`
`p_(SO_(2)) = (n_(SO_(2)))/(n_(T)) . p_("total")`
`:. (p_(SO_(3)))/(p_(O_(2))) = (n_(SO_(3)))/(n_(O_(2))) = (m_(SO_(3)))/(M_(SO_(3))) xx (M_(O_(2)))/(m_(O_(2)))`
Since `m_(SO_(3)) : m_(O_(2)) is 1 : 1` , we have
`:. (p_(SO_(3)))/(p_(O_(2))) = (M_(O_(2)))/(M_(SO_(3))) = (32 g mol^(-1))/(80 g mol^(-1)) = (2)/(5)`