At `1127 K` and `1 atm` pressure, a gaseous mixture of `CO` and `CO_(2)` in equilibrium with solid carbon has `90.55% CO` by mass:
`C_((s))+CO_(2(g))hArr2CO_((g))`
Calculate `K_(c)` for the reaction at the above temperature.

Explanation

If total mass of the mixture of `CO` and `CO_(2)` is `100 g`, then
`CO=90.55 g` and `CO_(2)=100-90.55=9.45 g`
`:.` Number of moles of `CO=90.55//28=3.234`
Number of moles of `CO_(2)=9.45//44=0.215`
`:. P_(CO)=3.234/(3.234+0.215)xx1 atm =0.938 atm`
`p_(CO_(2))=0.215/(3.234+0.215)xx1 atm =0.062 atm`
`K_(p)=p_(CO)^(2)/p_(CO_(2))=((0.938)^(2))/0.062=14.19`
`Deltan_(g)=2-1 :. K_(p)=K_(c )(RT)` or `K_(c )=K_(p)/(RT)`
`=14.19/(0.0821xx1127)=0.153`