Answer
$19.56 kg/m^{3}$
Work Step by Step
Given:
concentration of Si in Fe-Si alloy - 0.25 wt%
Required:
concentration in kg of Si per cubic meter alloy
Solution:
Using Equation 4.9a:
$c''_{Si} = \frac{c_{Si}}{\frac{c_{Si}}{p_{Si}}+\frac{c_{Fe}}{p_{Fe}}} \times 10^{3}$
where $p_{Si} = 2.33 g/cm^{3}$, $p_{Fe} = 7.87 g/cm^{3}$
Substituting the given values:
$c''_{Si} = \frac{0.25 wt}{\frac{0.25 wt}{2.33 g/cm^{3}}+\frac{99.75 wt}{7.87 g/cm^{3}}} \times 10^{3} = 19.56 kg/m^{3}$
