Question #435fb

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.