= Solution
Convert atomic fractions to <mass fractions> before using the diagram's weight-percent scale:
$$
w(x)=\frac{xM_{\rm Sn}}{xM_{\rm Sn}+(1-x)M_{\rm Bi}}.
$$
With $M_{\rm Sn}=118.71$ and $M_{\rm Bi}=208.98\,\mathrm{g\,mol^{-1}}$, the overall $x=0.30$ is \b[$19.58\,\mathrm{wt\%}$ Sn]. At $150\,{}^\circ\mathrm C$, the diagram gives approximately $x_B=0.028$ and $x_L=0.555$. These are about $1.61$ and $41.47\,\mathrm{wt\%}$ Sn. The <lever rule> gives
$$
\boxed{f_L=\frac{0.1958-0.0161}{0.4147-0.0161}\simeq0.45,\qquad f_B\simeq0.55.}
$$
The microstructure has primary Bi-rich crystals surrounded by remaining <liquid>. The phase compositions and fractions are approximate readings of a printed phase diagram, not exact thermodynamic values. Faster cooling generally gives finer crystals, solute segregation and departure from these equilibrium compositions.
Back to article page
