A titanium cube contains 2.86 * 10^23 atoms. What is the edge length l of the cube? The density of titanium is "4.50 g/cm"^3 (The volume of a cube is V=l^3)

Jan 18, 2018

$\text{1.72 cm}$

Explanation:

Start by using Avogadro's constant to find the number of moles of titanium that are present in the cube.

2.86 * 10^(23) color(red)(cancel(color(black)("atoms Ti"))) * overbrace("1 mole Ti"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ti")))))^(color(blue)("Avogadro's constant")) = "0.4749 moles Ti"

Next, use the molar mass of titanium to convert the number of moles to grams.

0.4749 color(red)(cancel(color(black)("moles Ti"))) * "47.867 g"/(1color(red)(cancel(color(black)("mole Ti")))) = "22.732 g"

Now, you know that ${\text{1 cm}}^{3}$ of titanium has a mass of $\text{4.50 g}$, so use the density of the metal to determine the volume of the cube.

22.732 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(4.50color(red)(cancel(color(black)("g")))) = "5.0516 cm"^3

Finally, use the fact that the length of a cube is equal to the cube root of its volume

$V = {l}^{2} \implies l = \sqrt[3]{V}$

to find the length of the cube.

l = root(3)("5.0516 cm"^3) = color(darkgreen)(ul(color(black)("1.72 cm")))

The answer is rounded to three sig figs.