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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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