# Chemistry - solid state

(i) Calculate its density in g cm−3 given that a = 4.203 Å.
Molecular weight of MgO : M = (24.3 + 16.0) g = 40.3 g mol−1
Vcell = a ^3 = 4.203 3 Å^3= 74.25 Å3
Convert to cm3
: Vcell = 74.25 ×(10−8)^3 cm3 = 74.25 ×10^−24 cm3
Convert to molar volume : V = (74.25 × 10^−24 × 6.022×1023) / 4 cm3 mol−1
= 11.18 cm3 mol−1
(since 4 MgO per unit cell) ----> this part I'm confused about
Density = M / V = 40.3 g mol−1 / 11.18 cm3
mol−1
= 3.605 g cm−3

How do you tell how many MgO units are in the unit cell?
Original post by JordanTRichie
(i) Calculate its density in g cm−3 given that a = 4.203 Å.
Molecular weight of MgO : M = (24.3 + 16.0) g = 40.3 g mol−1
Vcell = a ^3 = 4.203 3 Å^3= 74.25 Å3
Convert to cm3
: Vcell = 74.25 ×(10−8)^3 cm3 = 74.25 ×10^−24 cm3
Convert to molar volume : V = (74.25 × 10^−24 × 6.022×1023) / 4 cm3 mol−1
= 11.18 cm3 mol−1
(since 4 MgO per unit cell) ----> this part I'm confused about
Density = M / V = 40.3 g mol−1 / 11.18 cm3
mol−1
= 3.605 g cm−3

How do you tell how many MgO units are in the unit cell?

This explains it better than I could ...

https://chem.libretexts.org/Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_201_-_General_Chemistry_I_(Anthony_and_Clark)/Unit_4%3A_Quantifying_Chemicals/4.1%3A_Unit_Cells