Question

Palladium crystallizes with a face-contered cubic structure. It has a density of 12.0 g/cm". a radius of 138 pm, and a molar mass of 106.42 g/mol. vant to reference (Pages 508-516) Section 12.3 while completing this problem. Part these data to estimate Avogadro's number AEP OE? atom/mol 98

Answer 1

Ans- ince density is given by (d) 2х м. NAX V V- Volume - 93 2- number of ał om atoms per unit cell 2-4 ( for face- Centered Cubic) a = edge length , M = Molar Mass (Fed) NA = avog addo's number Now, for face-Cented Cubic (F cc) relation between radius and edge length is given by - radius (8)= √2 a . put value (8=138 pm) or 8 = 1.38 x108 cm so, a= 4 x 1-38x10-8 a= 3.90x108cm Now, put values in density foomula d=2x M Naxa 3 (2=4 (for Fcc)

12 =4X 106.42 NAX (3,90 X10-833 NA= 35.47 (3.90 X 10-833 NA= 5.98% 1023 or NA ñ 6.0 x 1023 avegado's number = 5.98 x 1023

Answer 2

Answer given densityd 12 g/cm3 radius (9) = 138 pmm. Molar mass of Palladium = 106.42 gimol. for face centre cubic . no. at alams pero unit cell te z =4 . for face centre cubic cell length a = 282r -.272 X 138x10 cm 3.90 X 10 cm. -23 a on volume (y) = 6.90x10 83=5193x102 cm3 we know the densiety ҳ м Where . M = vo NA VXNA Molar mass of Palladium, volume. = Avogadro Number: - NA = 2 xd 4x 106.42 unit calentation 5093X1023x129 5,98% 1023 mott cm3 gyan's g/mol a moit so nA = [sigs x 1023 atem/mol. Scanned with CamScanner