An element crystallizes in a face centered cubic lattice and has a density of 1.45 g cm-3 . The edge of its unit cell is 4.52 x 10-8cm.a) How many atoms are in each unit cell?
b) What is the volume of a unit cell?
c) What is the mass of a unit cell?
d) Calculate the approximate atomic mass of the element.

Answers

Answer 1

Answer:

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

The face-centered cubic lattice has 3 atoms from the 6 faces, and 1 atom from the eight corners. Thus, the total atoms in the face-centered lattice are four.

The face-centered lattice has been a cube.

The volume of cube = $\rm (edge)^3$

The volume of unit cell = $\rm (4.52\;*\;10^-^8\;cm)$

The volume of unit cell = $\rm 9.23\;*\;10^-^2^3\;cm^3$

The mass of a unit cell can be calculated from density. Mass can be defined as the ratio of volume to density.

Mass = $\rm (volume)/(density)$

Mass of unit cell = $\rm (9.23\;*\;10^-^2^3\;cm^3)/(1.45\;g\;cm^-^3)$

Mass of unit cell = $\rm 1.34\;*\;10^-^2^2\;g$ .

The approximate atomic mass of the element can be calculated by the mass of the carbon atom.

Mass of 1 carbon atom = $\rm (mass\;of\;1\;mole\;carbon)/(number\;of\;atoms\;in\;1\;mole\;Carbon)$

Mass of 1 carbon atom = $\rm (12)/(6.023\;*\;10^2^3)$

Mass of 1 carbon atom = 1.992 $\rm *\;10^-^2^3$ grams.

atomic mass unit per gram can be given as;

amu/gram = $\rm (12)/(1.992\;*\;10^-^2^3)$

amu/gram = $\rm 6.022\;*\;10^2^3$ amu/gram

1 gram = $\rm 6.022\;*\;10^2^3$ amu

1 amu = 1.661 $\rm *\;10^-^2^4$ gram.

The average atomic mass = mass of unit cell $*$ amu\gram

= $\rm 1.34\;*\;10^-^2^2\;g$ . $*$ 1 amu/ 1.661 $\rm *\;10^-^2^4$ gram.

= 80.7 amu.

In the given fcc element, a. the number of atoms is 4. b. The volume of a unit cell is $\rm 9.23\;*\;10^-^2^3\;cm^3$ . c. Mass of unit cell is $\rm 1.34\;*\;10^-^2^2\;g$ . d. The approximate atomic mass of the element is 80.7 amu.

For more information about the face-centered cubic lattice, refer to the link:

brainly.com/question/14578576

Answer 2

Answer:

Final answer:

A face centered cubic lattice consists of 4 atoms. The volume of the unit cell is 9.22 x 10^-23 cm^3 and the mass is 1.34 x 10^-23 g. The approximate atomic mass of the element is 2.02 amu.

Explanation:

The element is said to crystallize in a face centered cubic lattice. This implies that there is one atom at each corner of the cube (8 corners for a total of 1 atom, since each corner atom is shared among 8 adjacent cubes). There is also one atom on each face of the cube (6 faces for a total of 3 atoms, since each face atom is shared among 2 adjacent cubes). Thus, a total of 4 atoms are present in each unit cell. (a)

The volume of a unit cell (edges for a cube) can be calculated by the formula 'volume = side^3', where side in this case is 4.52 x 10-8cm. Hence, the volume equals (4.52 x 10^-8cm)^3 = 9.22 x 10^-23cm^3. (b)

The density of the substance is given as 1.45g/cm^3. The formula for density is 'mass/volume' which implies that mass can be calculated as 'density x volume'. Hence, the mass of the unit cell is (1.45g/cm^3) x (9.22 x 10^-23 cm^3) = 1.34 x 10^-23 g.(c)

The atomic weight of the element can then be calculated by taking this overall mass and dividing by the number of atoms in a unit cell (4). So, the atomic weight is (1.34 x 10^-23 g) / 4 = 3.35 x 10^-24 g. But atomic weights are usually given in atomic mass units (amu), not grams, and 1 amu = 1.66 x 10^-24 g. Therefore, we have an atomic weight of (3.35 x 10^-24 g) / (1.66 x 10^-24 g/amu) = approximately 2.02 amu. (d)

Learn more about Atomic Properties here:

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