Knowing that $[Ni(CO)_4]$ $=$ $0.85$ M at equilibrium for the reaction:
$Ni(s)+4CO(g)$ $Ni(CO)_4(g)$, with $K=5.0\cdot10^4 $ $M^{-1}$
it is asked to calculate the concentration in $\frac{g}{L}$ of $CO$ at equilibrium.
$K$ $=$ $\frac{[Ni(CO)_4]}{[CO]^4}$
$5.0\cdot10^4$ $=$ $\frac{0.85}{[CO]^4}$
$[CO]^4$ $=$ $\frac{0.85}{5.0\cdot10^4}$
$[CO]$ $=$ $(\frac{0.85}{5.0\cdot10^4})^{\frac{1}{4}}$ $=$ $0.0642$ M
In $1$ L, there are
$0.0642\;mol$ $=$ $0.0642\cdot M_{CO}$ $=$ $0.0642\cdot 28$ $=$ $1.8\;0g$
$c_{CO}$ $=$ $1.80 \frac{g}{L}$
$Ni(CO)_4(g)$, with $K=5.0\cdot10^4 $ $M^{-1}$
it is asked to calculate the concentration in $\frac{g}{L}$ of $CO$ at equilibrium.