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
Stefan V.
Jan 18, 2018

#"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 #"1 cm"^3# of titanium has a mass of #"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 = 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.

Related questions
See all questions in Density
Impact of this question
34233 views around the world
You can reuse this answer
Creative Commons License