Fast ! the density of aluminum is 2.70 g/cm3. how large a cube, in cm3, would contain 2.00 x 10^24 atoms of aluminum? use dimensional analysis to solve and show all work including units on every number

33.2 cm³

We are given;

Density of Aluminium is 2.70 g/cm³

Number of atoms of Aluminium is 2.00 × 10^24 atoms

We are required to calculate the volume of the cube;

We know that 1 mole of a compound or an element contains a number of particles equivalent to the Avogadro's constant.

Therefore;

1 mole of Aluminium contains 6.022 × 10^23 atoms

But, 1 mole of Al = 27 g

Therefore;

27 g of Al contains 6.022 × 10^23 atoms

We need to calculate the mass that contains 2.00 × 10^24 atoms

Moles of Al  = (2.00 × 10^24 atoms) ÷ (6.022 × 10^23 atoms)

                   = 3.32 moles  

Mass of Al = 27 g/mol × 3.32 moles

                 = 89.64 g

We can then calculate volume of Aluminium

Density = Mass ÷ Volume

Rearranging the formula;

Volume = Mass ÷ Density

            = 89.64 g ÷ 2.70 g/cm³

           = 33.2 cm³

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

i have no clue hope you find the answer