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A solution having composition $p$ (left side of graph) is cooled to just above the eutectic temperature (point $s$ is at $0.18$ $x_{\ce{Si}}$ and the corresponding tie line intersects the liquidus curve at $0.31$ $x_{\ce{Si}} $); calculate the composition of the solid that separates and that of the liquid that remains.

The liquid phase is obviously 31 % silicon and 69 % gold as given in the question. The solutions manual also gives the weight percentage of the solid phase. Using the lever rule,

$$ m_\text{solid}(0.18)=m_\text{liquid}(0.31-0.18)$$ $$ \frac{m_\text{solid}}{m_\text{liquid}}=\frac{0.13}{0.18} $$ $$ \frac{m_\text{solid}}{m_\text{solid}+m_\text{liquid}} = \frac{0.13}{0.13+0.18}=0.42$$

However, the solution manual states that the wt% of solid phase is 58 %, and that of the liquid phase is 42 %. Is the solutions manual wrong, or am I losing my mind?


You are correct. Another way to calculate it is: $$X_{\mathrm{L}} = \frac{s-b}{e-b} = \frac{0.18-0}{0.31-0}=0.58$$ You can also inspect the diagram visually and see that the $bs$ line is longer than the $se$ line. The lever rule dictates that the line length is proportional to the percentage of the phase opposite from the line. Thus, it is not possible that there is more solid than liquid at this point.

There is definitely a mistake in your textbook's solutions.


Using the lever rule, the amount of liquid in the 2 phase region is given by $$\frac{18-0}{31-0}=0.58$$

Hence for the amount of solid in the same region we get $$1-0.58=0.42$$

since the overall sum of the liquid and solid in the two phase region is 1.


There is no mistake in the solutions of your book. I hope to help any other students that may come across this post. The formula for the weight percentage of the solid phase is $$\text{wt% of solid} = \frac{\text{wt% of solid}}{\text{wt% of solid + wt% of liquid}}.$$

Applying this formula you get: $$\frac{0.18}{0.13 + 0.18}= 0.5806$$

Thus the $\text{wt% of solid}$ is $58.06\%$.

It is a common mistake to get the values backwards since it appears to be backwards from what it should be. I urge you to look up the formula for yourself to verify as well as check out a derivation of this formula.

  • 1
    $\begingroup$ @AvnishKabaj Please don't use \mathrm where it is neither necessary, nor correct. $\endgroup$ – Martin - マーチン Mar 19 '18 at 8:26
  • $\begingroup$ @Martin Isn't \mathrm just a font $\endgroup$ – Avyansh Katiyar Mar 19 '18 at 8:27
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    $\begingroup$ @AvnishKabaj No. It requests that symbols will be set in roman (upright) type, which might be correct, but it might be wrong, i.e. like here. Here it has no effect at all, as \text is always set in roman type. If you want to know more, please have a look here and here. $\endgroup$ – Martin - マーチン Mar 19 '18 at 8:32

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