An ornament weighs 36 g in air and 34 g in water. Assuming the ornament is prepared by mixing some copper with gold, determine the amount of copper present in it. The specific gravity of gold is 19.3 and that of copper is 8.9 .

Step-by-step solution

Loss of weight of the ornament in water = 36 - 34 = 2 g Since the density of water is 1 g/cm ^3 , the volume of the displaced water is 2 cm ^3 . Therefore, the total volume of the ornament is V = 2 cm ^3 . Let the mass of copper in the ornament be m . Then the mass of gold is 36 - m . The volume of copper is V_c = m 8.9 and the volume of gold is V_g = 36 - m 19.3 . The total volume of the ornament is the sum of the volumes of copper and gold: m 8.9 + 36 - m 19.3 = 2 m 8.9 - m 19.3 = 2 - 36 19.3 m ( 19.3 - 8.9 8.9 1