Question

A bag contains a mixture of copper and lead BBs. The average density of the BBs is 9.60g/cm3.Assuming that the copper and lead are pure, determine the relative amounts of each kind of BB.
Express your answers numerically separated by a comma.

Answer 1

Explanation:

Let the percentage of copper by weight present is x. Hence, then the percentage of lead will be (100 - x).

So, total mass = 100 g

Therefore, volume occupied by copper will be as follows.

                 Volume = $\frac{mass}{density}$

                               = $\frac{x}{8.94}$

Volume occupied by lead will be as follows.

                 Volume = $\frac{mass}{density}$

                               = $\frac{(100 - x)}{11.34}$

Total density = $\frac{\text{total mass}}{\text{total volume}}$

        9.60 $g/cm^{3}$ =  $\frac{100}{\frac{x}{8.94} + \frac{100 - x}{11.34}}$

                     x = 50.34

Therefore, percentage of lead will be (100 - 50.34) = 49.66.

Thus, we can conclude that percentage of copper is 50.34% and percentage of lead is 49.66%.

Answer 2

First, from reliable sources we get the densities of both the copper and lead. This will give us 8.96 g/cm3 and that of lead is 11.34 g/cm3. We assume that 1 cm3 of this certain mixture is composed of x cm3 of copper and y cm3 of zinc. Such that,
                                           x + y = 1
and                                   9.60 = x(8.96) + y(11.34)
The values of x and y are 0.73 cm3 and 0.27 cm3. The masses are obtained by multiplying these values by the densities giving us with 6.54 g and 3.06 g.