Question

An ornament weighing 36 g in air, weighs only 34 g 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 1

Given that,
$m_1$ a = 36 gm

$m_2$ water = 34 gm

Density of gold, ${\rho }_{Au}$ = 19.3 gm/cc

Density of copper, ${\rho }_{Cu}$ = 8.9 gm/cc

We know that, loss of wt = wt of displaced water

= 36-34 = 2 gm

Here, $m_e$ = $m_{Au}+m_{Cu}$

= 36 gm ...(i)

Let v be the volume of the ornament in cm.

So, $V\times {\rho }_w\times g=2\times g$

$\Rightarrow (v_{Au}+VCu)\times \rho w\times g=2\times g$

$\Rightarrow (\frac{m}{{\rho }_{Au}}+\frac{m}{{\rho }_{Cu}})\rho w\times g=2\times g$

$\Rightarrow (\frac{m_{Au}}{19.3}+\frac{m_{Cu}}{8.9})\times 1=2$

$\Rightarrow 8.9m_{Au}+19.3m_{Cu}=2\times 19.3\times 8.9$

= 343.54 ...(ii)

From eq. (i) and (ii),
8.9 $(m_{Au}+m_{Cu})=8.9\times 36$
i.e 8.9 $m_{Au}+8.9m_{Cu}$ = 320.40 ...(iii)
From (ii) & (iii), we get,
10.4 $m_{Cu}$ = 23.14
$\therefore m_{Cu}$= 2.225 gm

So, the amount of copper in the ornament is 2.2 g.