Understanding the Volume of a Large Number of Aluminum Atoms

Chemistry and physics often involve calculating the volume of substances based on the number of atoms and their properties. For example, given 3.00 × 1020 atoms of aluminum, you might need to determine the volume of the aluminum block these atoms would form. This process involves understanding the density of aluminum and applying basic chemical and physical principles.

Setting Up the Problem

First, let's establish the key parameters and values:

The density of aluminum, d 2.70 g/cm3 The total number of aluminum atoms, N 3.00 × 1020 The Avogadro number, NA 6.023 × 1023 atoms/mol The atomic mass of aluminum, MA 27 g/mol

Using these values, we can proceed to calculate the volume of the aluminum block.

Step-by-Step Calculation

1. **Convert the number of atoms to moles:** [ n frac{N}{N_A} frac{3.00 × 10^{20}}{6.023 × 10^{23}} approx 5.00 × 10^{-4} text{ mol} ]2. **Convert moles to grams:** [ m n × M_A 5.00 × 10^{-4} text{ mol} × 27 text{ g/mol} 1.35 × 10^{-2} text{ g} 0.0135 text{ g} ]3. **Use the density to find the volume:** [ V frac{m}{d} frac{0.0135 text{ g}}{2.70 text{ g/cm}^3} 0.005 text{ cm}^3 ]

Therefore, the volume of the aluminum block formed from 3.00 × 1020 atoms is 0.005 cm3.

Alternative Calculation for Verification

Alternatively, you can directly calculate the mass first and then determine the volume. Here's how:

Calculate the mass of 3.00 × 1020 atoms of aluminum in grams: Use the molar mass of aluminum, which is 27 g/mol, and the Avogadro number, which is 6.023 × 1023 atoms/mol: The mass of 3.00 × 1020 atoms can be calculated as follows:

Multiply the number of atoms by the atomic mass per atom divided by the Avogadro number:

[text{Mass} frac{3.00 × 10^{20} text{ atoms} × 27 text{ g/mol}}{6.023 × 10^{23} text{ atoms/mol}} approx 13.5 × 10^{-4} text{ g} 0.0135 text{ g} ]4. **Calculate the volume using the density:** [ V frac{m}{d} frac{0.0135 text{ g}}{2.70 text{ g/cm}^3} 0.005 text{ cm}^3 ]5. **Convert the volume to practical dimensions:** This volume can be visualized as a 'block' that is 1 mm × 1 mm × 5 mm, which is equivalent to 0.001 cm × 0.001 cm × 5 cm. This aligns with the calculation above.

Practical Application and Tips

Understanding these calculations is essential in various practical applications, such as in metallurgy, material science, and engineering. Here are some tips for accurately performing similar calculations:

Write all units: Always write down the units as you perform calculations. This helps in identifying and converting units properly. Check for unit cancellation: Ensure that the units you are dealing with cancel out appropriately, leaving the desired unit. Use significant figures: Be mindful of the significant figures in your calculations to maintain accuracy. Convert units as needed: Convert between grams, moles, and volume as required for the specific problem.

By following these steps and tips, you can accurately calculate the volume of a large number of atoms or other similar problems.

Conclusion

In summary, the process of calculating the volume of a large number of aluminum atoms involves converting the number of atoms to moles, then to grams, and finally using the density to determine the volume. Understanding and applying these principles ensures accurate results and enhances your problem-solving skills in chemistry and physics.