Question

Determine the approximate density of a high-leaded brass that has a composition of 61.5 wt% Cu, 33.5 wt% Zn, and 5.0 wt% Pb. The densities of Cu, Zn, and Pb are 8.94, 7.13 and 11.35 g/cm3, respectively.

Answer 1

To solve this problem it is necessary to apply the concept related to the definition of Density. Density is defined as the relationship between volume and mass. The greater the volume with respect to the mass, the smaller the mass tends to be and vice versa. By definition it is given by the formula:

$\rho = \frac{m}{V}$

Where,

V= Volume

m = Mass

From the data given we have to,

$\% of Cu = 61\% \Rightarrow \rho_{Cu}=8.96g/cm^3$

$\% of Zn = 33.5\%\Rightarrow \rho_{Zn}= 7.14g/cm^3$

$\% of Pb = 5\%\Rightarrow \rho_{Pb}=11.35g/cm^3$

1g of the alloy will contain 0.61g Cu, 0.335g Zn and 0.05g Pb

Therefore if we calculate the volume of each element we have,

$V=\frac{m}{\rho}$

$V_{Cu}=\frac{0.61g}{8.96g/cm^3}=0.06808cm^3$

$V_{Zn}=\frac{0.335g}{7.14g/cm^3}=0.0469cm^3$

$V_{Pb}=\frac{0.05g}{11.35g/cm^3}=0.0044cm^3$

$V_{total} = 0.73221cm^3$

The density of the alloy would be

$\rho = \frac{m}{V}$

$\rho = \frac{1g}{0.73221}$

$\rho = 1.3659g/cm^3$

Therefore the density of the alloy is $1.3659g/cm^3$