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Question: A bcc element (atomic mass 65) has a cell edge of 420 pm. Calculate its density in \(g/c{m^{ - 3}}\)...

A bcc element (atomic mass 65) has a cell edge of 420 pm. Calculate its density in g/cm−3g/c{m^{ - 3}}.

Explanation

Solution

Density is the measure of mass per unit of volume of a substance. It is denoted by the symbol ρ\rho (the greek letter rho). The average density of a substance can be calculated by its total mass divided by its total volume. Substances made from a dense material such as an iron will have less volume than the substances made from some less dense substance such as water. The density of a substance varies with the temperature and pressure of the substance. The variation is small for solids and liquids but greater for gases. When the pressure of the substance increases due to which the volume of the substance decreases and thus its density increases.

Complete step by step answer:
In the given question the atomic mass and the edge length of a bcc (Body-centered cubic) element are given and using these given values we have to calculate the density of the bcc element.
Given,
Atomic mass (M) of the element =65 = 65
Edge length (a) =420pm = 420pm
=420×10−10cm= 420 \times {10^{ - 10}}cm , (1pm=10−10cm)(1pm = {10^{ - 10}}cm)
=42×10×10−10cm= 42 \times 10 \times {10^{ - 10}}cm
=42×10−9cm= 42 \times {10^{ - 9}}cm
For an bcc element, Z=2Z = 2
Now using these given values and applying the density formula we will calculate the density of the bcc element,
ρ=M×ZNA×a3\rho = \dfrac{{M \times Z}}{{{N_A} \times {a^3}}}
where NA={N_A} = Avogadro’s number
⇒ρ=65×26.023×1023×(42×10−9)3\Rightarrow \rho = \dfrac{{65 \times 2}}{{6.023 \times {{10}^{23}} \times {{\left( {42 \times {{10}^{ - 9}}} \right)}^3}}}
⇒ρ=1306.023×1023×(42)3×10−27\Rightarrow \rho = \dfrac{{130}}{{6.023 \times {{10}^{23}} \times {{\left( {42} \right)}^3} \times {{10}^{ - 27}}}}
⇒ρ=1306.023×74088×10−4\Rightarrow \rho = \dfrac{{130}}{{6.023 \times 74088 \times {{10}^{ - 4}}}}
⇒ρ=130446232.024×104\Rightarrow \rho = \dfrac{{130}}{{446232.024}} \times {10^4}
⇒ρ=0.000291×104g/cm−3\Rightarrow \rho = 0.000291 \times {10^4}g/c{m^{ - 3}}
⇒ρ=2.91 g/cm−3\Rightarrow \rho = 2.91\,g/c{m^{ - 3}}

So,the density of the bcc element is found to be 2.91 g/cm−3g/c{m^{ - 3}}.

Note: The Cubic system is the simplest. There are three common types of lattices depending upon the unit cells. These are:
Simple Cubic Lattice: - In this type of lattice, there are lattice points only at the eight corners of the cube. Each corner is shared between eight adjacent unit cells.
Body-Centered Cubic Lattice: - This type of lattice has pointed at the corners of a cube with an additional point at the center of the cube.
Face Centered Cubic Lattice: - In this type of lattice, there are points at the corners of the cube with additional points at the center of each face.