The concentration of \(\mathrm{Si}\) in an Fe-Si alloy is \(0.25 \mathrm{wt} \%\). What is the concentration in kilograms of Si per cubic meter of allov?

We know that the density of an alloy (ρ_alloy) is given by the formula: $$ρ_{alloy} = \frac{m_{alloy}}{V}$$ where \(m_{alloy}\) is the mass of the alloy and \(V\) is the volume of the alloy. Since we don't have the density of the alloy given in the problem, we are going to assume the density of the alloy is the weighted average density of Si and Fe, which is a good approximation for many alloys. So, the density of Fe-Si alloy is: $$ρ_{alloy} = ρ_{Fe} × wt\%_{Fe} + ρ_{Si} × wt\%_{Si}$$ The densities of pure Fe and Si are approximately \(ρ_{Fe} = 7874 \frac{kg}{m^3}\) and \(ρ_{Si} = 2330 \frac{kg}{m^3}\), respectively. The given weight percentage of Si is \(0.25\%\) and therefore, the weight percentage of Fe is \(100\% - 0.25\% = 99.75\%\).

To find the mass of Si in a unit volume of the alloy, we can use the given weight percentage of Si: $$m_{Si} = m_{alloy} × wt\%_{Si}$$

Using the mass and density of the alloy, we can find the volume: $$V = \frac{m_{alloy}}{ρ_{alloy}}$$

Now, we can find the concentration of Si in the alloy (in kg/m³) using the mass of Si and the volume of the alloy: $$C_{Si} = \frac{m_{Si}}{V}$$ Combining Steps 1-4, we get: $$C_{Si} = \frac{m_{alloy} × wt\%_{Si}}{\frac{m_{alloy}}{ρ_{Fe} × wt\%_{Fe} + ρ_{Si} × wt\%_{Si}}}$$ $$C_{Si} = \frac{wt\%_{Si} × (ρ_{Fe} × wt\%_{Fe} + ρ_{Si} × wt\%_{Si})}{wt\%_{Fe}}$$ Plugging in the values for \(wt\%_{Si}\), \(wt\%_{Fe}\), \(ρ_{Fe}\) and \(ρ_{Si}\): $$C_{Si} = \frac{0.0025 × (7874 × 0.9975 + 2330 × 0.0025)}{0.9975}$$

Now we can calculate the final concentration of Si in kg/m³: $$C_{Si} = \frac{0.0025 × (7858.95 + 5.825)}{0.9975} ≈ 2336.33 \frac{kg}{m^3}$$ So, the concentration of Si in the Fe-Si alloy is approximately \(2336.33 \frac{kg}{m^3}\).