At `727^(@)C` and `1.2 atm` of total equilibrium pressure, `SO_(3)` is partially dissociated into `SO_(2)` and `O_(2)` as:
`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)`
The density of equilibrium mixture is `0.9 g//L`. The degree of dissociation is:, `[Use R=0.08 atm L mol^(-1) K^(-1)]`

Options

A. `1//3`
B. `2//3`
C. `1//4`
D. `1//5`

Answer

B

Explanation

`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g) M_(mix)=(0.9xx0.082xx1000)/(1.23)=60`
`1-alpha alpha (alpha)/(2) M_(mix)=(alphaM_(SO_(2))+(alpha)/(2)M_(O_(2))+(1-alpha)M_(SO_(3)))/(1+(alpha)/(2))`
`1+(alpha)/(2)=(80)/(60) (alpha)/(2)=(20)/(60)`
`alpha=(2)/(3)`