= Solution
$S=e/e_s^l$ is the ambient <saturation ratio> over a flat liquid-<water> surface; $S-1$ is the liquid <atmospheric supersaturation>. Curvature increases the surface equilibrium vapour <pressure>, producing the <Kelvin equation>[Kelvin penalty] $A/r$, while dissolved solute lowers it and gives the opposing <Raoult's law>[Raoult contribution] $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\rm eq}-1=A/r-B/r^3$, and the numerator is the <atmospheric supersaturation> available to drive growth beyond surface equilibrium.
$F_d^l$ describes the resistance of vapour <diffusion> through the surrounding air. $F_k^l$ describes the <thermal conduction>[thermal-conduction] resistance to removing <latent heat> released by <condensation>: a warmed droplet surface has a higher equilibrium vapour <pressure> and hence a weaker <mass>-transfer drive. Common <continuum approximations> are $F_d^l\simeq R_vT/(D_ve_s^l)$ and $F_k^l\simeq L_v[k_{\rm air}T]^{-1}[L_v/(R_vT)-1]$, showing the roles of <diffusivity>, conductivity, <saturation pressure> and <latent heat>. Both resistances have the supplied units $\mathrm{s\,m\,kg^{-1}}$ and add for this coupled transport problem.
For a spherical droplet of constant <density>, $m=4\pi\rho r^3/3$. Multiplying the <radius> equation by $4\pi\rho r$ gives
$$
\frac{dm}{dt}=\frac{4\pi r}{F_k^l+F_d^l}
\left[(S-1)-\frac Ar+\frac B{r^3}\right].
$$
At sufficiently large <radius>, curvature and solute corrections are negligible, so
$$
\boxed{\frac{dm}{dt}\simeq4\pi r\frac{S-1}{F_k^l+F_d^l}.}
$$
This limit also requires sustained positive <atmospheric supersaturation>. It need not describe the earliest submicrometre stage or an unactivated haze particle.
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