Answers:
(a) [tex]0.0073kg[/tex]
(b) Volume gold: [tex]3.79(10)^{-7}m^{3}[/tex], Volume cupper: [tex]7.6(10)^{-8}m^{3}[/tex]
(c) [tex]17633.554kg/m^{3}[/tex]
Explanation:
We are told the total mass [tex]M[/tex] of the coin, which is an alloy of gold and copper is:
[tex]M=m_{gold}+m_{copper}=7.988g=0.007988kg[/tex] (1)
Where [tex]m_{gold}[/tex] is the mass of gold and [tex]m_{copper}[/tex] is the mass of copper.
In addition we know it is a 22-karat gold and the relation between the number of karats [tex]K[/tex] and mass is:
[tex]K=24\frac{m_{gold}}{M}[/tex] (2)
Finding [tex]{m_{gold}[/tex]:
[tex]m_{gold}=\frac{22}{24}M[/tex] (3)
[tex]m_{gold}=\frac{22}{24}(0.007988kg)[/tex] (4)
[tex]m_{gold}=0.0073kg[/tex] (5) This is the mass of gold in the coin
The density [tex]\rho[/tex] of an object is given by:
[tex]\rho=\frac{mass}{volume}[/tex]
If we want to find the volume, this expression changes to: [tex]volume=\frac{mass}{\rho}[/tex]
For gold, its volume [tex]V_{gold}[/tex] will be a relation between its mass [tex]m_{gold}[/tex] (found in (5)) and its density [tex]\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}[/tex]:
[tex]V_{gold}=\frac{m_{gold}}{\rho_{gold}}[/tex] (6)
[tex]V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}[/tex] (7)
[tex]V_{gold}=3.79(10)^{-7}m^{3}[/tex] (8) Volume of gold in the coin
For copper, its volume [tex]V_{copper}[/tex] will be a relation between its mass [tex]m_{copper}[/tex] and its density [tex]\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}[/tex]:
[tex]V_{copper}=\frac{m_{copper}}{\rho_{copper}}[/tex] (9)
The mass of copper can be found by isolating [tex]m_{copper}[/tex] from (1):
[tex]M=m_{gold}+m_{copper}[/tex]
[tex]m_{copper}=M-m_{gold}[/tex] (10)
Knowing the mass of gold found in (5):
[tex]m_{copper}=0.007988kg-0.0073kg=0.000688kg[/tex] (11)
Now we can find the volume of copper:
[tex]V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}[/tex] (12)
[tex]V_{copper}=7.6(10)^{-8}m^{3}[/tex] (13) Volume of copper in the coin
Remembering density is a relation between mass and volume, in the case of the coin the density [tex]\rho_{coin[/tex] will be a relation between its total mass [tex]M[/tex] and its total volume [tex]V[/tex]:
[tex]\rho_{coin}=\frac{M}{V}[/tex] (14)
Knowing the total volume of the coin is:
[tex]V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}[/tex] (15)
[tex]\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}[/tex] (16)
Finally:
[tex]\rho_{coin}=17633.554kg/m^{3}}[/tex] (17) This is the total density of the British sovereign coin
