= Solution
Let $\Gamma$ be coverage in $\mathrm{mol\,m^{-2}}$. Then $n_s=N_A\Gamma$ and the Helmholtz surface-dipole relation is
$$
\Delta\Phi_{\rm eV}=-\frac{N_A\Gamma p_\perp}{\epsilon_0},
$$
where a voltage in volts has the same numerical value as the electron-energy change in <electronvolts>. Converting the slope from square centimeters to square meters gives $-5\times10^5\ \mathrm{V\,m^2\,mol^{-1}}$. Therefore
$$
\boxed{p_\perp=\frac{\epsilon_0(5\times10^5)}{N_A}
=7.35\times10^{-30}\ \mathrm{C\,m}=2.20\ \mathrm D.}
$$
\b[Each added <molecule> produces an effective outward surface dipole of about $2.2$ D], lowering the <electron> <work function>. This is the incremental adsorbate-plus-surface dipole, not necessarily the isolated molecular <electric dipole moment>.
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