An ornament weighting 36g in air, weighs only 34g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9.

Answer

B

Explanation

Given that `m_1a=36gm`
`m_2 water =34gm`
Density of gold `rho_(An)=19.3gm//cc`
Density of copper, `rho_(Cu)=8.9gm/cc`
we know that loss of `wt=wt of displaced water`
`=36-34=2gm`
Here, `m_c=m_(Au)+m_(Cu)`
`=36gm`…….i
Let v be the volume of the orN/Ament in cm.
S, `Vxxrho_wxx=2xxg`
`rarr (v_(Au)+V_(Cu))xxrho_wxxg=2xxg`
`rarr (m/(rho_(Au))+m/(rho_(Cu)))rho_wxxg=2xxg`
`rarr ((m_(Au))/19.3+(m_(Cu))/8.9)x1=2`
`rarr 8.9m_(Au)+19.3m_(Cu)=2xx19.3x8.9`
`=343.54`..........ii
Form eq i and ii ltbr. `8.9(m_(Au)+m_(Cu))=8.9xx36`
i. `8.9m_(Au)+8.9m_(Cu)=320.40`.........iii ltbr. `from ii and iii we get ,
`10.4m_(Cu)=23.14`
`=m_(Cu)=2.225gm`
so the amount of copper in the oerN/Ament is 2.2g.