A pure titanium cube has an edge length of 2.66 in. how many titanium atoms does it contain? titanium has a density of 4.50 g/cm3.

The volume of the cube will be 2.66 * 2.66 * 2.66 = 18.821 cubic inches.

1 cubic inch = 16.387 cubic centimeter.

18.821 cubic inches will be 18.821 * 16.387 = 308.42 cubic centimeter.

Density is 4.5 g/cm³

Therefore, mass will be 308.42 * 4.5 = 1387.89 g

1 mole of Titanium = 47.877 g, so the number of moles will be 1387.89/47.877=28.99 moles.

1 mole = 6.02 * 10^23 atoms

Therefore, 28.99 moles * 6.02 * 10^23 atoms = 1.745 * 10^25 atoms.

Volume of Cube = edge length^3

V=2.66*2.66*2.66

V=18.82 in^3

because density is in cm^3

so converting V from in^3 to cm^3

V=314.12 cm^3

mass=density * volume

mass = 4.5*314.12

mass=1413.54 g

The molar mass of titanium is 47.88 g/mole

so moles=1413.54/47.88

moles=29.52

now one mole has avogadro number of atoms

so number of atoms=29.52*avogadro number

number of atoms=1.77x10^25