Atitanium 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