Question

46.3 wt%A, 23.15 wt%B and 30.55 wt%C. Assume the densities in g/cu.cm of A, B andC are 2.68, 5.86 and 10.39, respectively. Determine the approximate density in g/cu.cm if the alloy is composed of

Answer 1

Answer : The approximate density is
`5.77g/cm^3`

Explanation :

The formula used to calculate the density of the mixture is,

`\rho=[(m_A* \rho_A)+(m_B* \rho_B)+(m_C* \rho_C)]`

where,

`\rho` = density of mixture = ?

`m_A` = mass of fraction of component A = 46.3 % = 0.463

`m_B` = mass of fraction of component B = 23.15 % = 0.2315

`m_C` = mass of fraction of component C = 30.55 % = 0.3055

`\rho_A` = density of component A =
`2.68g/cm^3`

`\rho_B` = density of component B =
`5.86g/cm^3`

`\rho_C` = density of component C =
`10.39g/cm^3`

Now put all the given values in the above formula, we get:

`\rho=[(0.463* 2.68)+(0.2315* 5.86)+(0.3055* 10.39)]`

`\rho=5.77g/cm^3`

Therefore, the approximate density is
`5.77g/cm^3`