Question

8. All of the alkali metals adopt the same solid structure-a body-centered cubic unit cell. The molar mass of lithium is 6.94 g/cm. The length of an edge of its unit cell is 3.507 Å. The molar mass of cesium is 132.91 g/cm; its unit cell edge length is 6.147 Å. a. What is the radius for each of these atoms? b. What is the volume of space (in Å) that is unoccupied by atoms (i.e., amount of empty space in the unit cell?) c. What is the % empty space in each of these unit cells?

Answer 1

stay for a body centered mit cell 4 ur Voa. a mandues of in as edge length: For lithium, a = 3.507Å . ar B x 3.507 4 u = 1.518. resium , a= 6:1474 r= √3 x 6. 147Ă - For 2.66Ã . 43 (CN=2). .03 3.507 A. by volume of unit cell = a3 = Volume of ions occupied in unit cell = 2x 4 43 For lithium, mit rcella a 13 - )3 A Volume of ions occupied - 2*4. x 3.14x17:51) N92288943 Holume of empty space = 43.13 43 – 28.82 Å = 14.30 Å? Volu (Co call) Volume of Unit cella a² = (6. 1478) 3 - 232.2683 Holume of ions occupied 328 / ar² = 2X 4% * 3. 14x(2.66) AS z 157, 5983 Holume of empty space = (232.26 - 187.59) Å3 , 14 - 6 6 8.

7.1. empty space - space a Volume of empty space Volume of empty volume of unit call x 100%. For lithium, 7. empty space = 14.30A x 100%. 43. 13 A03 = 33.15) For Cesium, space = 76.66 7. empty Ås 232.26 A3 X 100 % = 32.147.