# Question 435fb

Feb 28, 2017

Here's what I got.

#### Explanation:

The idea here is that you can express the density of a substance as the ratio between the mass of a given sample of said substance, let's say $m$, and the volume it occupies, let's say $V$.

color(blue)(ul(color(black)(rho = m/V)))" " " "color(red)("(*)")

Now, let's take ${m}_{\text{Au}}$ to be the mass of gold and ${m}_{\text{Cu}}$ the mass of copper used to make the mixture.

You know that

${m}_{\text{Au" + m_"Cu" = "668 g}}$

If you take ${V}_{\text{Au}}$ to be the volume of gold and ${V}_{\text{Cu}}$ to be the volume of copper in the mixture, you can rewrite the above equation by using equation $\textcolor{red}{\text{(*)}}$

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

This means that you have

${\rho}_{\text{Au" * V_"Au" + rho_"Cu" * V_"Cu" = "668 g"" " " }} \textcolor{b l u e}{\left(1\right)}$

Here

• ${\rho}_{\text{Au}}$ is the density of gold, equal to ${\text{19.3 g cm}}^{- 3}$
• ${\rho}_{\text{Cu}}$ is the density of copper, equal to ${\text{8.9 g cm}}^{- 3}$

You also know that

${V}_{\text{Au" + V_"Cu" = "40 cm"^3 " " " }} \textcolor{b l u e}{\left(2\right)}$

You now have two equations with two unknowns. Rewrite equation $\textcolor{b l u e}{\left(2\right)}$ as

${V}_{\text{Au" = "40 cm"^3 - V_"Cu}}$

and plug in into equation $\textcolor{b l u e}{\left(1\right)}$ to find

$19.3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g")))"cm"^(-3) * ("40 cm"^3 - V_"Cu") + 8.9 color(red)(cancel(color(black)("g"))) "cm"^(-3) * V_"Cu" = 668 color(red)(cancel(color(black)("g}}}}$

This is equivalent to

19.3 color(red)(cancel(color(black)("cm"^(-3)))) * 40 color(red)(cancel(color(black)("cm"^3))) - "19.3 cm"^(-3) * V_"Cu" + "8.9 cm"^(-3) * V_"Cu" = 668

which gives

$772 - \text{10.4 cm"^(-3) * V_"Cu} = 668$

and thus

V_"Cu" = (772 - 668)/"10.4 cm"^(-3) = color(darkgreen)(ul(color(black)("10 cm"^3)))

Plug this into equation $\textcolor{b l u e}{\left(2\right)}$ to find the value of ${V}_{\text{Au}}$

V_"Au" = "40 cm"^3 - "10 cm"^3 = color(darkgreen)(ul(color(black)("30 cm"^3)))#

Therefore, you can say that the mixture was made by mixing ${\text{30 cm}}^{3}$ of gold and ${\text{10 cm}}^{3}$ of copper.