Question

If the atomic radius of a simple cubic metal is 0.126 nm and its atomic weight is 70.4 g/mol 1. What is the length of its unit cell in nanometers (nm) and centimeters (cm)? 2. What is the volume of its unit cell in cubic nanometers and cubic centimeters? 3. What is the density of the metal in grams per cubic centimeter (cc)?

Answer 1

Answer:

1. $a = 0.252nm$

$a=2.52*10^{-8}cm$

2. $V=0.016nm^{3}$

$V=1.6*10^{-23}cc$

3. $d=14.6\frac{g}{cc}$

Explanation:

1. Find the length of the unit cell:

a. In nanometers (nm):

$a=2R$

$a=2*0.126nm$

$a=0.252nm$

b. Convert the radius from nm to cm to calculate the length of the unit cell In centimeters (cm):

$0.126nm*\frac{1*10^{-7}cm}{1nm}=1.26^{-8}cm$

c. Apply the formula to find the lentht of the unit cell in cm:

$a=2R$

$a=2*(1.26*10^{-8})$

$a=2.52*10^{-8}cm$

2. Find the volume of the unit cell:

$V=a^{3}$

- In nanometers (nm):

$V=(0.252nm)^{3}$

$V=0.016nm^{3}$

- In cubic centimeters (cc):

$V=(2.52*10^{-8}cm)^{3}$

$V=1.6*10^{-23}cc$

3. Find the Density of the metal in grams per cubic centimeter:

$d=\frac{(atoms/cell)*atomicweight}{unitcellvolume*Avogadrosnumber}$

$d=\frac{(2\frac{atoms}{cell})*(70.4\frac{g}{mol})}{(1.6*10^{-23}cc)*(6.022*10^{23}\frac{atoms}{mol})}$

$d=14.6\frac{g}{cc}$