Question #435fb

1 Answer
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_"Au" to be the mass of gold and m_"Cu" the mass of copper used to make the mixture.

You know that

m_"Au" + m_"Cu" = "668 g"

If you take V_"Au" to be the volume of gold and V_"Cu" to be the volume of copper in the mixture, you can rewrite the above equation by using equation color(red)("(*)")

rho = m/V implies m = rho * V

This means that you have

rho_"Au" * V_"Au" + rho_"Cu" * V_"Cu" = "668 g"" " " "color(blue)((1))

Here

  • rho_"Au" is the density of gold, equal to "19.3 g cm"^(-3)
  • rho_"Cu" is the density of copper, equal to "8.9 g cm"^(-3)

You also know that

V_"Au" + V_"Cu" = "40 cm"^3 " " " " color(blue)((2))

You now have two equations with two unknowns. Rewrite equation color(blue)((2)) as

V_"Au" = "40 cm"^3 - V_"Cu"

and plug in into equation color(blue)((1)) to find

19.3 color(red)(cancel(color(black)("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 - "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 color(blue)((2)) to find the value of V_"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 "30 cm"^3 of gold and "10 cm"^3 of copper.