Heterogeneous chemical equilibrium

We know that $[Ni(CO)_4]$ $=$ $0.85$ M at the equilibrium: $Ni(s)$ $+$ $4CO(g)$ $Ni(CO)_4(g)$, with $K$ $=$ $5.0\cdot10^4 $ $M^{-1}$ So: $K = $ ..........$\frac{[Ni(CO)_4]}{[CO]^4}$ from where we get $[CO]$: ..........$5.0\cdot10^4$ $=$ $\frac{0.85}{[CO]^4}$
$[CO]^4$ $=$ $\frac{0.85}{5,0\cdot10^4}$
$[CO]$ $=$ $(\frac{0.85}{5,0\cdot10^4})^{\frac{1}{4}}$ $=$ $0.0642$ M Concentration c expressed in $\frac{g}{L}$: Mass of $CO$ in $1$ L = ..........$0.0642\;mol$ $ =$ $0.0642\cdot M_{CO}$ $=$ $0.0642\cdot 28$ $=$ $1,80\;g$ $c_{CO}$ $=$ ..........$1,80\frac{g}{L}$