Barium has a bcc unit cell with a length of 508pm along an edge.what i

1.	Barium has a bcc unit cell with a length of 508pm along an edge.what is the density of barium in g cm^3
Answer» ANSWER: edge length of unit cell = 508 pm so, volume of unit cell = (508pm)³ = 1.31096512 × 10^(-36 + 8) m³ ≈ 1.3 × 10^-28 m³ a/c to question, barium has a body centered cubic unit cell. so NUMBER of barium atom in a unit cell = 2. so, volume of a barium atom = volume of unit cell/2 = 1.3 × 10^-28/2 m³ = 0.65 × 10^-28 m³ volume of one mole of barium atoms, V = 6.022 × 10²³ × 0.65 × 10^-28 m³ = 3.9143 × 10^-5 m³ [ as we KNOW, 1 m³ = 10^6 cm³ ] = 39.143 cm³ atomic mass of barium = 137.327 g so, density of barium = atomic mass of barium/volume of one mole of barium atom = 137.327/39.143 g/cm³ ≈ 3.51 g/cm³ plz mark me brainliest

Barium has a bcc unit cell with a length of 508pm along an edge.what is the density of barium in g cm^3

Answer»

edge length of unit cell = 508 pm

so, volume of unit cell = (508pm)³

= 1.31096512 × 10^(-36 + 8) m³

≈ 1.3 × 10^-28 m³

a/c to question, barium has a body centered cubic unit cell.

so NUMBER of barium atom in a unit cell = 2.

so, volume of a barium atom = volume of unit cell/2 = 1.3 × 10^-28/2 m³

= 0.65 × 10^-28 m³

volume of one mole of barium atoms, V = 6.022 × 10²³ × 0.65 × 10^-28 m³

= 3.9143 × 10^-5 m³

[ as we KNOW, 1 m³ = 10^6 cm³ ]

= 39.143 cm³

atomic mass of barium = 137.327 g

so, density of barium = atomic mass of barium/volume of one mole of barium atom

= 137.327/39.143 g/cm³

≈ 3.51 g/cm³

plz mark me brainliest