Aufgaben:
a) \(\displaystyle \rho=\frac12\cdot\rho_\text{Au}+\frac12\cdot\rho_\text{Ag}=\frac12\cdot19,32\frac g{cm^3}+\frac12\cdot10,49\frac g{cm^3}=14,9\frac g{cm^3}\)
b) \(\displaystyle \rho=1/\sum \limits_{i}\frac{m_{\%,i}}{\rho_i}=1/\left(\frac1{2\cdot19,32\frac g{cm^3}}+\frac1{2\cdot10,49\frac g{cm^3}}\right)=13,6\frac g{cm^3}\)
c)
\(\displaystyle (i) \rho=0,58\cdot0,789\frac g{cm^3}+0,42\cdot1\frac g{cm^3}=0,878\frac g{cm^3}\)
\(\displaystyle (ii) m_{\%,\text{H}_2\text{O}}=\frac{0,42\cdot1\frac g{cm^3}}{0,878\frac g{cm^3}}=47,84\%\)
d) \(\displaystyle x\cdot19,32\frac g{cm^3}+y\cdot10,49\frac g{cm^3}=15\frac g{cm^3} \land x+y=1\)
\(\displaystyle\quad\Rightarrow x=0,51\land y=0,49\)
1. Rechnung über die Volumenanteile:
\(\displaystyle\quad V_{\text{Leg}}=\frac m\rho=\frac{1kg}{15000\frac{kg}{m^3}}=66,67cm^3\)
\(\displaystyle\quad V_\text{Au}=0,51\cdot66,67cm^3=34cm^3\)
\(\displaystyle\quad m_\text{Au}=V_\text{Au}\cdot\rho_\text{Au}=34cm^3\cdot 19,32\frac g{cm^3}=656,88g\)
2. Möglichkeit mit den Massenanteilen:
\(\displaystyle\quad m_{\%,\text{Au}}=\frac{0,51\cdot19,32\frac g{cm^3}}{15\frac g{cm^3}}=65,688\%\)
\(\displaystyle\quad 1kg\cdot0,65688=656,88g\)
