Molybdenum forms a substitutional solid solution with tungsten. Compute the weight percent of molybdenum that must be added to tungsten to yield an alloy that contains \(1.0 \times\) \(10^{22}\) Mo atoms per cubic centimeter. The densities of pure Mo and \(\mathrm{W}\) are \(10.22\) and \(19.30\). \(\mathrm{g} / \mathrm{cm}^{3}\), respectively.

Given the concentration of Mo atoms in the alloy, we know that there are \(1.0 \times 10^{22}\) Mo atoms per cubic centimeter. To find the number of W atoms in the same volume, we need to use the densities of pure Mo and W. Let the volume fractions of Mo and W be \(V_{Mo}\) and \(V_{W}\), respectively, and their densities be \(\rho_{Mo}\) and \(\rho_{W}\). The total volume (1 cm³) is their sum: \(V_{Mo} + V_{W} = 1\). We can express the mass of Mo and W as \(m_{Mo} = \rho_{Mo} V_{Mo}\) and \(m_{W} = \rho_{W} V_{W}\), respectively. Now, let's use the Avogadro's number (\(N_A\)) to find the number of atoms of Mo and W in the alloy. Given that there are \(1.0 \times 10^{22}\) Mo atoms per cubic centimeter, we can get the number of W atoms in the alloy as follows: \(N_{W} = \frac{m_{W}}{M_{W}} N_{A}\), where \(M_{W}\) is the molar mass of W.

We can use the equation above to find the number of moles of W and Mo in the alloy: \(n_{W} = \frac{N_{W}}{N_{A}}\) and \(n_{Mo} = \frac{N_{Mo}}{N_{A}}\).

We can now calculate the weight of Mo and W in the alloy by multiplying the number of moles by their respective molar masses: \(m_{Mo} = n_{Mo} M_{Mo}\) and \(m_{W} = n_{W} M_{W}\)

Finally, we can find the weight percent of Mo in the alloy: Weight percent of Mo \(= \frac{m_{Mo}}{m_{Mo} + m_{W}} \times 100\) By calculating the weight percent of Mo in the alloy using the given densities and the concentration of Mo atoms, we can determine how much molybdenum must be added to tungsten to get the desired alloy concentration.