{"id":183565,"date":"2025-01-16T16:24:24","date_gmt":"2025-01-16T16:24:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183565"},"modified":"2025-01-16T16:24:26","modified_gmt":"2025-01-16T16:24:26","slug":"calculate-the-theoretical-density-of-al-from-the-following-data","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/calculate-the-theoretical-density-of-al-from-the-following-data\/","title":{"rendered":"Calculate the theoretical density of Al from the following data"},"content":{"rendered":"\n

Calculate the theoretical density of Al from the following data: Crystal Structure = FCC Atomic radius = 1.86 \u00c5 Atomic weight = 26.98 g\/mo<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To calculate the theoretical density of aluminum (Al) with a face-centered cubic (FCC) crystal structure, an atomic radius of 1.86 \u00c5, and an atomic weight of 26.98 g\/mol, we can follow these steps:<\/p>\n\n\n\n

1. Determine the Number of Atoms per Unit Cell (n):<\/strong><\/p>\n\n\n\n

In an FCC structure, there are 4 atoms per unit cell.<\/p>\n\n\n\n

2. Calculate the Edge Length (a) of the Unit Cell:<\/strong><\/p>\n\n\n\n

The relationship between the atomic radius (R) and the edge length (a) in an FCC structure is given by:<\/p>\n\n\n\n

[ a = 2\\sqrt{2} \\times R ]<\/p>\n\n\n\n

Given that the atomic radius R = 1.86 \u00c5 = 1.86 \u00d7 10\u207b\u00b9\u2070 m, we can calculate:<\/p>\n\n\n\n

[ a = 2\\sqrt{2} \\times 1.86 \\times 10^{-10} \\, \\text{m} ]<\/p>\n\n\n\n

[ a \\approx 4.24 \\times 10^{-10} \\, \\text{m} ]<\/p>\n\n\n\n

3. Calculate the Volume of the Unit Cell (V\u2080):<\/strong><\/p>\n\n\n\n

The volume of the unit cell is the cube of the edge length:<\/p>\n\n\n\n

[ V_0 = a^3 ]<\/p>\n\n\n\n

[ V_0 = (4.24 \\times 10^{-10} \\, \\text{m})^3 ]<\/p>\n\n\n\n

[ V_0 \\approx 7.64 \\times 10^{-29} \\, \\text{m}^3 ]<\/p>\n\n\n\n

4. Calculate the Mass of the Unit Cell (m\u2080):<\/strong><\/p>\n\n\n\n

The mass of the unit cell is the product of the number of atoms per unit cell (n), the atomic weight (M), and Avogadro’s number (N\u2090):<\/p>\n\n\n\n

[ m_0 = n \\times M \/ N_a ]<\/p>\n\n\n\n

[ m_0 = 4 \\times 26.98 \\, \\text{g\/mol} \/ 6.022 \\times 10^{23} \\, \\text{atoms\/mol} ]<\/p>\n\n\n\n

[ m_0 \\approx 1.79 \\times 10^{-22} \\, \\text{g} ]<\/p>\n\n\n\n

5. Calculate the Theoretical Density (\u03c1):<\/strong><\/p>\n\n\n\n

Density is mass per unit volume:<\/p>\n\n\n\n

[ \\rho = m_0 \/ V_0 ]<\/p>\n\n\n\n

[ \\rho = 1.79 \\times 10^{-22} \\, \\text{g} \/ 7.64 \\times 10^{-29} \\, \\text{m}^3 ]<\/p>\n\n\n\n

[ \\rho \\approx 2.34 \\times 10^{3} \\, \\text{g\/m}^3 ]<\/p>\n\n\n\n

[ \\rho \\approx 2.34 \\, \\text{g\/cm}^3 ]<\/p>\n\n\n\n

Therefore, the theoretical density of aluminum is approximately 2.34 g\/cm\u00b3.<\/p>\n\n\n\n

Explanation:<\/strong><\/p>\n\n\n\n

The theoretical density of a crystalline material can be calculated using the formula:<\/p>\n\n\n\n

[ \\rho = \\frac{n \\times M}{V_0 \\times N_a} ]<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n