# Question #2b165

##### 1 Answer

#### Explanation:

Your strategy for this problem will be to use the edge length of the unit cell to determine the unit cell's *volume*, then use tungsten's relative atomic mass to find the mass of a *single* tungsten atom.

Once you know that mass of a single tungsten atom, use the total number of atoms that can fit into a *body-centered cubic* lattice to find the mass of the unit cell.

Since density is usually given in *grams per cubic centimeters*, it will be useful to start by converting the edge length of the unit cell from *picometers* to *centimeters*.

#"1 cm" = 10^(-2)"m" = 10^(10)"pm"#

This means that you have

#316.5color(red)(cancel(color(black)("pm"))) * "1 cm"/(10^10color(red)(cancel(color(black)("pm")))) = 3.165 * 10^(-8)"cm"#

As you know, the volume of a cube is given by the formula

#color(blue)(V = l xx l xx l = l^3)" "# , where

In your case, the volume of the unit cell will be

#V = (3.165 * 10^(-8))^3 "cm"^3 = 3.170 * 10^(-23)"cm"^3#

Now, the *unified atomic mass unit*,

#"1 u" = 1.660539 * 10^(-27)"kg"#

This means that the mass of a *single* tungsten atom, expressed in **grams**, will be equal to

#183.8color(red)(cancel(color(black)("u"))) * (1.660539 * 10^(-27)color(red)(cancel(color(black)("kg"))))/(1color(red)(cancel(color(black)("u")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 3.052 * 10^(-22)"g"#

Now, a **body-centered cubic** lattice is characterized by a unit cell that contains a total of

Each *corner* lattice point contains *center* lattice point contains one atom. Since you have

#"no. of atoms" = overbrace(1/8 xx 8)^(color(blue)("8 corners")) + 1 = "2 atoms"#

This means that the **total mass** of a tungsten unit cell will be

#2color(red)(cancel(color(black)("atoms W"))) * (3.052 * 10^(-22)"g")/(1color(red)(cancel(color(black)("atom W")))) = 6.014 * 10^(-22)"g"#

Finally, density is defined as mass per unit of volume. In tungsten's case, the mass of the unit cell and its volume are characteristic of a density of

#color(blue)(rho = m/V)#

#rho = (6.014 * 10^(-22)"g")/(3.170 * 10^(-23)"cm"^3) = color(green)("19.25 g/cm"^3)#

The listed value for tungsten's density is *excellent result*