Question

Gold is alloyed(mixed) with other metals to increase its hardness in making jewelry. (A) Consider a piece of gold jewelry that weighs 9.85 g and has a volume of 0.675 cm3. The jewelry contains only gold and silver, which have densities of 19.3 g/cm3 and 10.5 g/cm3, repectively. If the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains. Calculate the percentage of gold(by mass) in the jewelry. (B) The relative amount of gold in an alloy is commonly expressed in units of karats.

Answer 1

To calculate the average density of the element, we take the summation of the product of the density of the pure element and the percent abundance. The density of the jewelry is 9.85 g/ 0.675 cm3 equal to 14.59 g/cm3. The equation is 14.59 = 19.3 * x+ 10.5*(1-x) where x is the percent abundance of pure gold. The percent abundance of gold is 46.78% and that of silver is 53.52%.

Answer 2

Answer:

The percentage of gold(by mass) in the jewelry is 61.52%.

Explanation:

Mass of the jewelry ,M= 9.85 g

Mass of the gold in jewelry = $m_1$

Mass of the silver in jewelry = $m_2$

Volume of the jewelry = $V=0.675 cm^3$

Volume of the gold in jewelry = $v_1$

Density of gold $d_1=19.3 g/cm^3=\frac{m_1}{v_1}$

$m_1=19.3g/cm^3 \times v_1$

Volume of the silver in jewelry = $v_2$

Density of gold $d_2=10.5 g/cm^3=\frac{m_2}{v_2}$

$m_2=10.5g/cm^3 \times v_2$

$V=v_1+v_2$

$v_1+v_2=0.675 cm^3$..(1)

$m_1+m_2=M$

$d_1v_1+d_2v_2= 9.85 g$

$(19.3 g/cm^3)v_1+(10.5 g/cm^3)v_2= 9.85 g$..(2)

On solving equation (1)and(2) we get:

$v_1=0.314 cm^3,v_2=0.361 cm^3$

and after that values of $m_1 andm_2$

$m_1=19.3g/cm^3 \times 0.314 cm^3=6.0602 g$

$m_2=10.5g/cm^3 \times 0.316 cm^3=3.7905 g$

The percentage of gold(by mass) in the jewelry:

$\frac{m_1}{M}\times 100$

$\frac{6.0602 g}{3.7905 g}\times 100=61.52\%$

The percentage of gold(by mass) in the jewelry is 61.52%.