= Solution
In the high-surface-potential <Schulze-Hardy rule>, the <critical coagulation concentration> scales approximately as $z^{-6}$ with the <counterion> valence. Hence
$$
c_{\rm Al}=c_{\rm Ca}\left(\frac{2}{3}\right)^6
=0.01\frac{64}{729}
=\boxed{8.78\times10^{-4}\ \mathrm{mol\,L^{-1}}}.
$$
The more highly charged <counterion> therefore requires about eleven times less <concentration> in this idealized estimate. This assumes the cations are the <counterions> of a negatively charged surface, the surface-potential regime is appropriate, and the only changing mechanism is electrostatic stabilization. Specific <adsorption> and hydrolysis of aluminium <ions> can invalidate this scaling; matching <ionic strengths> instead is not the same prediction.
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