Question

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 1

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