At 1127 K and 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))`
`K_(c)` for this reaction at the above temperature is

Explanation

If the total mass of the gaseous mixture at equilibrium is 100 g, then
mass of `CO(g) = 90.55 g `
and mass of `CO_2(g) = 100 - 90 . 55 = 9.45`g
` therefore ` Number of moles of CO (g) ` = (90 . 55)/(28) = 3.234`
` " " (because ` molar mass of CO= 28 molar mass of `CO_2=44)`
and number of moles of ` CO_2(g) = (9.45)/(44) = 0.215`
Total number of gaseous moles = ` 3.234 +0.215 = 3.449 `
` therefore " " rho_(CO) = (3.234)/(3.449) xx 1 = 0.938 " atm " `
and ` rho _(CO_2)= ( 0.215)/( 3.449) xx 1= 0.062 ` atm
` ( because ` partial pressure = ` (" no of moles ")/( " total no. of moles ") xx ` total pressure )
Hence, for the reactions.
` " " C( s) + CO_2(g) hArr 2CO_2(g)`
` K_(rho) = ( rho_(CO) ^(2))/( rho_(CO_2)) = ((0.938)^(2))/( 0.062) = 14.19 ` For the above reactions ` Delta n = 2-1=1`
` because " " K_rho = K_c ( RT)^(Delta n ) `
` " " K_c = (K_p)/( (RT)^(Delta n )) = ( 14.19 )/( (0.0821 xx 1127 )^(1)) = 0.153 `