Question

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

(a) Find the mass of gold in the sovereign in kilograms using the fact that the number of karats = 24× (mass of gold)/ total mass.

(b) Calculate the volumes of gold and copper, respectively, used to manufacture the coin.

(c) Calculate the density of the British sovereign coin.

Answer 1

a

The mass of gold is
`L = 7.322 *10^(-3) \ kg`

b

The volumes of gold and copper is
`V_g = 3.794 *10^(-7) \ m^3` ,
`V_c = 7.426 *10^(-8) \ m^3`

c

The density of the British sovereign coin

`\rho = 17.593*10^(3) \ kg/m^3`

Step-by-step explanation:

From the question we are told that

The total mass of the gold is
`K = 7.988 \ g = 7.988 * 10^(-3) \ kg`

The karat of the British gold sovereign is
`z = 22`

Let the mass of gold in the alloy be L

Now we are told that

`z = 24 * (L)/(K)`

substituting value

`22 = 24 * (L)/(7.988 * 10^(-3))`

So
`L = (22)/(24) * 7.899*10^(-3)`

`L = 7.322 *10^(-3) \ kg`

The volume of the gold coin is mathematically represented as

`V_g = (L)/(\rho_g )`

Where
`\rho_g` is the density of the gold which a constant with value

`\rho_g = 19.3 *10^(3) \ kg /m^3`

So

`V_g = (7.322 *10^(-3))/(19.3 *10^(3) )`

`V_g = 3.794 *10^(-7) \ m^3`

The mass of copper is mathematically evaluated as

`m_c = K - L`

`m_c = 7.988*10^(-3) - 7.322 *10^(-3)`

`m_c = 6.657 *10^(-4) \ kg`

Volume of the copper is

`V_c = (m_c)/(\rho_c)`

Where
`\rho_c` is the density of the copper which a constant with value

`\rho_c = 8.92 * 10^(3) \ kg/m^3`

So

`V_c = (6.657 *10^(-4))/(8.92 *10^(3))`

`V_c = 7.426 *10^(-8) \ m^3`

The total volume of the British gold sovereign coin is \

`V = V_g + V_c`

substituting values

`V = 3.7939 *10^(-7) + 7.4626 *10^(-7)`

`V = 4.54 *10^(-7) \ m^3`

The density of the British gold sovereign coin is

`\rho = (K)/(V)`

substituting values

`\rho = (7.988 *10^(-3))/(4.54 *10^(-7))`

`\rho = 17.593*10^(3) \ kg/m^3`

Answer 2

(a)
`m_(gold)=7.322g`

(b)

`V_(gold)=0.379cm^3`

`V_(copper)=0.122cm^3`

(c)
`\rho _(coin)=15.94g/cm^3`

Step-by-step explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

`m_(gold)=(m_(tota)*karats)/(24)=(7.988g*22)/(24)=7.322g`

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

`V_(gold)=(m_(gold))/(\rho_(gold)) =(7.322g)/(19.32g/cm^3)=0.379cm^3`

`m_(copper)=7.988g-7.322g=1.09g\\V_(copper)=(m_(copper))/(\rho_(copper))=(1.09g)/(8.94g/cm^3) \\\\V_(copper)=0.122cm^3`

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

`\rho _(coin)=(m_(total))/(V_(gold)+V_(copper))=(7.988 g)/(0.379cm^3+0.122cm^3)\\ \\\rho _(coin)=15.94g/cm^3`

Best regards.