= Solution
At two-phase equilibrium, the endpoints of a common tangent to the two molar free-energy curves give the coexisting compositions. They are not the separate curve minima. Reading the graph gives approximately $x_{B,\alpha}\simeq0.02$ and $x_{B,m}\simeq0.64$, with a few atomic-percent graphical uncertainty.
Let $f_\alpha$ be the fraction of atoms in the <precipitate> <thermodynamic phase>. Conservation of the overall B fraction gives
$$
0.55=f_\alpha x_{B,\alpha}+
(1-f_\alpha)x_{B,m},
$$
and hence
$$
\boxed{f_\alpha=\frac{x_{B,m}-0.55}
{x_{B,m}-x_{B,\alpha}}
\simeq\frac{0.64-0.55}{0.64-0.02}
\simeq0.145\approx15\%.}
$$
About85percent remains in the B-richer matrix. Smooth readings of the original curves put the result near14–15percent; the graph does not justify many significant figures. The result is an atomic or molar <thermodynamic phase> fraction. Its identification with a <volume fraction> in the next estimate requires comparable atomic volumes of the two <thermodynamic phases>.
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