At room temperature, sodium crystallizes in a body centred cubic lattice with \(a=4.24 \AA\). Calculate theoretical density of sodium (At. wt. of \(\mathrm{Na}=23\) ).

Short Answer

Expert verified
The theoretical density of sodium is found by calculating the mass of sodium atoms in a BCC unit cell and dividing it by the volume of the unit cell.

Step by step solution

01

Calculate the volume of the unit cell

First, determine the volume of the body-centered cubic (BCC) unit cell using the lattice parameter 'a'. The formula for the volume of a cube is given by the cube of the edge length. Hence, the volume of the unit cell (V) is calculated with the formula: \( V = a^3 \)
02

Determine the number of atoms per unit cell for BCC

In a body-centered cubic (BCC) lattice, there is 1 atom at each corner of the cube and 1 atom at the center. Each corner atom is shared by 8 adjacent unit cells, so each corner contributes 1/8 of an atom to the unit cell. Because there are 8 corners, the corner atoms contribute a total of 1 atom. The center atom belongs entirely to the unit cell. Therefore, the total number of atoms in a BCC unit cell is: \( 1 \text{ (from corners)} + 1 \text{ (from center)} = 2 \text{ atoms} \)
03

Calculate mass of sodium atoms in the unit cell

Knowing the atomic weight (At. wt.) of sodium (\text{Na}) and the number of atoms in a unit cell, we can calculate the mass of the atoms in the unit cell. The mass of an atom in atomic mass units (amu) is numerically equal to the atomic weight. The mass of a single \text{Na} atom in grams is therefore the atomic weight divided by Avogadro's number (\text{N}_A). Multiply this by the number of atoms per unit cell to get the total mass (m) for the cell: \( m = \frac{\text{Atomic weight of Na}}{\text{N}_A} \times \text{Number of atoms per unit cell} \)
04

Compute the theoretical density of sodium

Finally, the theoretical density (\rho) is calculated using the mass (m) of the atoms in the unit cell and the volume (V) of the unit cell. Use the formula for density: \( \rho = \frac{m}{V} \). Convert the mass to grams and the volume to cubic centimeters if needed to get the density in \text{g/cm}^3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Body-Centered Cubic Lattice
The body-centered cubic (BCC) lattice represents a common crystal structure found in many metals such as iron at certain temperatures. In this type of lattice, atoms are located at each corner of a cube and with one atom at the exact center of the cube. Imagining multiple BCC cells stacked together shows that the corner atoms are shared among eight unit cells, while the center atom is exclusive to its particular unit cell. This layout for the arrangement of atoms within the crystal affects many of the material's properties, including its density.

Because of sharing, when we calculate the total number of atoms within any single BCC unit cell, we must consider that each corner atom contributes only an eighth of an atom to the cell since the corners are shared by eight neighboring cells. Therefore, summing up the contribution from all eight corners and the singular center atom results in a total of two atoms per BCC unit cell.
Unit Cell Volume
The volume of a unit cell is the space it occupies, which is crucial when determining properties like density. For a cubic lattice, such as the BCC, calculating the volume is straightforward. Since a cube has all sides of equal length, known as the lattice parameter 'a,' the volume is simply the cube of this length, expressed as \( V = a^3 \).

In our case with a sodium (Na) unit cell with a given edge length of 4.24 angstroms, this calculation gives the volume of the space taken up by two sodium atoms in the BCC structure. Converting the volume to a more manageable unit like cubic centimeters can be necessary, especially when comparing materials or evaluating the practical application in manufacturing or engineering contexts.
Avogadro's Number
Avogadro's number, denoted by \( N_A \), is a fundamental constant in chemistry representing the number of particles, usually atoms or molecules, in one mole of a substance. The value of Avogadro's number is approximately \( 6.022 \times 10^{23} \) entities per mole. This incredibly large number helps in converting measurements made on the atomic scale to macroscopic scales that are more meaningful in everyday contexts.

For instance, when calculating the mass of atoms in a unit cell, as seen in the exercise, Avogadro's number enables us to determine the mass of a single atom or a mole of atoms in grams, linking the atomic weight scale to the gram scale. This is essential when we are interested in the amount of substance in terms of mass, which is a common scenario in material science, chemistry, and physics.
Atomic Weight
Atomic weight, also known as relative atomic mass, is a dimensionless physical quantity expressed in atomic mass units (amu). It represents the average mass of atoms of an element, taking into account the different isotopes and their abundances. For sodium (Na) specifically, the atomic weight is approximately 23 amu. This means that an average sodium atom is 23 times heavier than 1/12 of the mass of a carbon-12 atom.

The atomic weight helps in understanding the mass of individual atoms when considering their contribution to the overall mass of a compound or element in bulk form. When calculating theoretical densities, the atomic weight allows us to deduce the mass of atoms in a unit cell from the macroscopic mass that can be measured in laboratories. Furthermore, by combining atomic weight with Avogadro's number, we can convert atomic mass to grams, which is a standard unit for measuring mass in the scientific and everyday world.

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Most popular questions from this chapter

Which of the following reactions will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into the right or left direction? (i) \(\mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{~S}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CS}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{~S}_{(\mathrm{g})}\) (ii) \(\mathrm{CO}_{2(\mathrm{~g})}+\mathrm{C}_{(\mathrm{s})} \rightleftharpoons 2 \mathrm{CO}_{(\mathrm{g})}\) (iii) \(4 \mathrm{NH}_{3(\mathrm{~g})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NO}_{(\mathrm{g})}+6 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}\) (iv) \(\mathrm{C}_{2} \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6(\mathrm{~g})}\)

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