Volume displaced by equal-mass pure gold, $V_{Au}$
0.22 L
SG gold / silver
19.32 / 10.50
Find. The purity of the crown, % gold by weight, if the balance of its mass is silver.
Approach. The crown and the pure-gold reference have the SAME total mass $M$ (that is how Archimedes matched them on the balance beam), so $V_{Au}=M/\rho_{Au}$ fixes $M$ directly. The crown's larger displaced volume is then split between a gold fraction $x$ (by mass) and a silver fraction $(1-x)$, each contributing its own volume at its own density; solving for $x$ from the crown's total displaced volume gives the purity.
Mass of the crown (= mass of the pure-gold reference).
$$M=\rho_{Au}V_{Au}=(19{,}320)(0.22\times10^{-3})=\boxed{4.2504\ \text{kg}}$$
Volume balance for the gold–silver crown. With mass fraction gold $x$, the alloy's total displaced volume is
$$V_{crown}=xM/\rho_{Au}+(1-x)M/\rho_{Ag}=x\,V_{Au}+(1-x)\frac{M}{\rho_{Ag}}$$
Substituting $M/\rho_{Ag}=4.2504/10{,}500=4.048\times10^{-4}\ \text{m}^3$ and $V_{crown}=3.30\times10^{-4}\ \text{m}^3$,
$$3.30\times10^{-4}=x(2.20\times10^{-4})+(1-x)(4.048\times10^{-4})$$
Solve for the gold mass fraction.
$$3.30\times10^{-4}-4.048\times10^{-4}=x\!\left(2.20\times10^{-4}-4.048\times10^{-4}\right)\ \Rightarrow\ x=\frac{-0.748\times10^{-4}}{-1.848\times10^{-4}}=0.4048$$
$$\boxed{\%\,\text{gold by weight}=40.5\%}$$
