Materingchemisty Homework B-Materials Google onome Secure https//session 8182165
ID: 512530 • Letter: M
Question
Materingchemisty Homework B-Materials Google onome Secure https//session 81821650 Engineering Chemistry l Signed in as lynx enve vancol t Unit Cells and Density t Unit Cells and Density Formula for density The unt cel of a crystal can be desaited n tem Based on he number of aoms per unit oell and the mass of the atom, mass mof the unt cer can be of edge length Land the number of The edge length defines the volume of a calculated The density of the unit cel and the materialas a whole can be determined from the mass mand the volume Vor the une cel as cubic unt cell, V, as density The usual unts of density are games per cubic centimeter or g/em' The number or Momi per unit depends on the type of unt cel For cubic unit cels the posnon (body face or comer) of the atoms detemines the Part A tahon of the atoms that are completely contaned within the unt cell Nickel Ni has a face centered outic 1)wth anedge length of 352 pm What is the densry ot this metan Use the petiodictate as noeded Express your answer with the appropriate units. densty Value Units MAnswers Give up Review part ContinueExplanation / Answer
Since in FCC the atoms occupy corners aswellas at the centers of six faces, FCC has 4 atoms per unit cell.
It can be calculated as follows,
each corner atom is shared by eight unitcells,
so contribution by corner atom per unitcell = 8 * (1/8) = 1
Each face center atom is shared by 2 unitcells,
SO, contribution by face center atom per unit cell = 2*(1/2) = 1
Therefore,
total number of atoms in FCC unit cell = 1 (from 8 corners) + 3 ( from 6 face centers) = 4 atoms
Molar mass of Ni = 58.7 g/mol
Therefore,
mass of 6.022 * 1023 atoms of Ni = 58.7 g.
then, mass of 4 atoms of Ni = 4 * 58.7 / (6.022 * 1023) = 3.90 * 10-22 g.
Volume = l3 = (352 pm)3 = (352 * 10-12)3 m3 =4.36*10-29m3=4.36 * 10-29 * 106 cm3 = 4.36*10-23cm3
Therefore,
density = mass / volume
d = 3.90 * 10-22 / (4.36*10-23)
d = 8.94 g/cm3