Question

determine the density of CsCl which crystallizes in a bc type structure with edge length 412.1pm​

Answer 1

Answer:strcuture of Cesium chloride is body centre cubic type in which Cs+ placed within centre and Cl- placed at corner of lattice.

number of Cl- ions per unit cell = 8 × 1/8 = 1

number of Cs+ ions per unit cell = 1

so, there is only one molecule of CsCl in a lattice .

so, mass of one molecule of CsCl = molar mass of CsCl/Avogadro's number

= (133 + 35.5)/(6.023 × 10²³)

= 168.5/6.023 × 10²³

= 2.8 × 10^-22 g

volume of lattice = a³ , where a is edge length

= (412.1 × 10^-12)³

= 7 × 10^-23 cm³

so, density of CsCl crystal = 2.8 × 10^-22/7 × 10^-23

= 4 g/cm³