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.
$C_{(s)} + CO_{2(g)} \longleftrightarrow 2CO_{(g)}$
Calculate $K_c$ for this reaction at the above temperature.

(N/A) Let the total mass of the gaseous mixture be $100 \, g$.
Mass of $CO = 90.55 \, g$
Mass of $CO_2 = 100 - 90.55 = 9.45 \, g$
Moles of $CO$,$n_{CO} = \frac{90.55}{28} = 3.234 \, mol$
Moles of $CO_2$,$n_{CO_2} = \frac{9.45}{44} = 0.215 \, mol$
Total moles = $3.234 + 0.215 = 3.449 \, mol$
Partial pressure of $CO$,$p_{CO} = \frac{3.234}{3.449} \times 1 = 0.938 \, atm$
Partial pressure of $CO_2$,$p_{CO_2} = \frac{0.215}{3.449} \times 1 = 0.062 \, atm$
$K_P = \frac{(p_{CO})^2}{p_{CO_2}} = \frac{(0.938)^2}{0.062} = 14.19$
For the reaction,$\Delta n = 2 - 1 = 1$
Using $K_P = K_C(RT)^{\Delta n}$:
$14.19 = K_C(0.0821 \times 1127)^1$
$K_C = \frac{14.19}{92.5257} \approx 0.153$.