A titanium cube contains 2.86*1023 atoms. what is the edge length l of the cube? the density of titanium is 4.50 g/cm3. (the volume of a cube is v=l3.)

When mass Ti = density * volume

and when moles Ti = mass Ti/molar mass Ti

∴ Volume = 2.86 x 10^23 atom * (1 mol Ti / 6.022 x 10^23) * (47.867 g Ti / 1 mol Ti) * (1Cm3 / 4.5 g Ti)

= 5.05 Cm^3

when we assume that the sample of Ti is a cube:

and we assume the length = X

∴ V = X^3

∴X^3 = 5.05

∴X = ∛5.05

= 1.7 Cm