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.

Answer 1

Answer:

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

          $\rho = 17.593*10^{3} \ kg/m^3$

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

       $z = 24 * \frac{L}{K}$

substituting value  

        $22 = 24 * \frac{L}{7.988 * 10^{-3}}$

  So  $L = \frac{22}{24} * 7.899*10^{-3}$

        $L = 7.322 *10^{-3} \ kg$

The volume of the gold coin is  mathematically represented as

         $V_g = \frac{L}{\rho_g }$

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

     $\rho_g = 19.3 *10^{3} \ kg /m^3$

So

         $V_g = \frac{7.322 *10^{-3}}{19.3 *10^{3} }$

         $V_g = 3.794 *10^{-7} \ m^3$

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  

          $V_c = \frac{m_c}{\rho_c}$

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

        $\rho_c = 8.92 * 10^{3} \ kg/m^3$

So

       $V_c = \frac{6.657 *10^{-4}}{8.92 *10^{3}}$

       $V_c = 7.426 *10^{-8} \ m^3$

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

      $\rho = \frac{K}{V}$

substituting values

         $\rho = \frac{7.988 *10^{-3}}{4.54 *10^{-7}}$

         $\rho = 17.593*10^{3} \ kg/m^3$

Answer 2

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

(c) $\rho _{coin}=15.94g/cm^3$

Explanation:

Hello,

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

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(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:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$

$m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3$

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

$\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3$

Best regards.