= Solution
Convert the new <standard concentration> to the number-density convention:
$$
1\ \mathrm{mol\,dm^{-3}}=10^3N_{\mathrm A}\ \mathrm{m^{-3}}
=6.02214076\times10^{26}\ \mathrm{m^{-3}}.
$$
For this association, $K_c=\rho_{\mathrm{H_2CO}}c^\circ/(\rho_{\mathrm H_2}\rho_{\mathrm{CO}})$, so it is proportional to $c^\circ$. Therefore
$$
\boxed{K_c^{(1\,\mathrm{mol\,dm^{-3}})}
=(5.2026\times10^{-32})(6.02214\times10^{26})
\simeq3.13\times10^{-5}.}
$$
Both dimensionless constants describe the same physical equilibrium; changing the reference changes the numerical activity quotient because two reactant molecules form one product molecule.
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