# 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#)

##### 1 Answer

#### 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 **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 = root(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**.