An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is 2.77 , and the density of the crystal is 8.65 . Find the atomic weight of the element.

55.34g/mol^3

a=2.77 Å= 2.77•10⁻⁸ cm.

The BCC-cell volume is
V=a³=(2.77•10⁻⁸)³ = 2.13•10⁻²³ cm³.
Density ρ =8.65g/cm³ per unit cell.
Avogadro´s number N₀=6.022•10²³ mol⁻¹.
The number of atoms in the BCC unit cell are calculated
as follows:
1 × 1 = 1 center atom
8 × (1/8) = 1 corner atom
=> n=2 total atoms

Atomic weight = N₀•V• ρ/n =
=6.022•10²³•2.13•10⁻²³•8.65/2 = 55.34 g/mol³

To find the atomic weight of the element, we need to follow these steps:

Step 1: Determine the volume of the unit cell.
Given that the edge of the unit cell is 2.77 Å, which is equivalent to 2.77 x 10^-10 meters (1 meter = 10^10Å), we can calculate the volume of the unit cell.
The volume of a cube can be found by raising the length of one side to the power of 3.
Volume of the unit cell = (Edge length)^3
Volume of the unit cell = (2.77 x 10^-10 meters)^3

Step 2: Convert the volume to cubic meters.
To get the volume in cubic meters, we need to multiply the volume of the unit cell by the conversion factor (1 Å^3 = 10^-30 m^3).
Volume of the unit cell = (2.77 x 10^-10 meters)^3 x (1 Å^3 / 10^-30 m^3)

Step 3: Calculate the mass of the unit cell.
The density of the crystal, which is given as 8.65 g/cm^3, is equivalent to 8.65 x 10^3 kg/m^3 (since 1 g/cm^3 = 10^3 kg/m^3).
Mass of the unit cell = Density x Volume of the unit cell

Step 4: Convert the mass to atomic mass units (amu).
Since the mass given is in kilograms (kg), we need to convert it to atomic mass units (amu) using the conversion factor (1 kg = 6.022 x 10^26 amu).
Atomic weight of the element = Mass of the unit cell x (6.022 x 10^26 amu / 1 kg)

By following the above steps and plugging in the given values, the atomic weight of the element can be calculated.

To find the atomic weight of the element, we need to use the given information about the unit cell edge length and the density of the crystal.

First, let's determine the volume of the unit cell. In a body-centered cubic (BCC) lattice, there is one whole atom at each of the eight corners of the cube, plus one additional atom in the center of the cube. So, there are a total of two atoms per unit cell.

The volume of a cube can be calculated by raising the length of one side to the power of three, which in this case is the edge length of the unit cell. Therefore, the volume of the unit cell in cubic meters (m³) is:

Volume of unit cell = (2.77 m)^3

Next, we need to convert the given density from grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³). Since 1 g/cm³ is equal to 1000 kg/m³, the density of the crystal in kg/m³ is:

Density of crystal = 8.65 g/cm³ × 1000 kg/m³

Now, let's calculate the mass of the crystal in the unit cell using the equation:

Mass = Density × Volume

The mass is equal to the density multiplied by the volume:

Mass = (8.65 kg/m³) × (2.77 m)^3

Now, we know that the mass is equal to the atomic weight (in grams) multiplied by Avogadro's number (6.022 × 10^23), divided by the molar mass (in grams/mole):

Mass = Atomic weight × Avogadro's number / Molar mass

Since we want to find the atomic weight, we can rearrange the equation as follows:

Atomic weight = (Mass × Molar mass) / Avogadro's number

By substituting the values we have obtained, we can calculate the atomic weight.

Please note that the molar mass depends on the specific element, so we would need additional information or context in order to determine it.