The radius of silver atom is 143.5 pm and it crystallises in face centred cubic arrangement. (molecular mass of `Ag = 107.87, N_A = 6.023 xx 10^(23)` ) What is the edge length of the unit cell?
A
405.8 pm
B
40.6 pm
C
331.4 pm
D
287.0 pm
Text Solution
Verified by Experts
Topper's Solved these Questions
QUESTION PAPER 2022 TERM 1
ICSE|Exercise question|1 Videos
PROLEMS BASED ON CHEMICAL STRUCTURES AND REACTIONS
ICSE|Exercise QUESTIONS|185 Videos
SAMPLE PAPER 3 (CHEMISTRY)
ICSE|Exercise QUESTIONS|70 Videos
Similar Questions
Explore conceptually related problems
The density of a face centred cubic element (atomic mass = 40 ) is 4.25 gm cm^(-3) , calculate the edge length of the unit cell.
The radius of copper atom is 128 pm. It it crystallises in face centred cubic lattice (fcc), what will be the length the edge of the unit cell ?
The radius of chromium atom is 1.25Å . If it crystallises in body centred cubic lattice, calculate the length of the edge of the unit cell.
Calculate the density of silver which crystallizes in face- centred cubic form. The distance between nearest metal atoms is 287 pm (Molar mass of Ag = 107.87 g mol^(-1) , N_A = 6.022 xx 10^(23) mol^(-1) ).
A metal X crystallises in a face-centred cubic arrangement with the edge length 862 pm. What is the shortest separation of any two nuclei of the atom ?
An element (at. mass = 60) having face centred cubic unit cell has a density of 6.23g cm^(-3) . What is the edge length of the unit cell? (Avogadro's constant = 6.023 xx 10^(23) mol^(-1) )
Gold (atomic mass = 197 u) has atomic radius = 0.144 nm. It crystallises in face centred unit cell. Calculate the density of gold. (No = 6.022xx10^(23)mol^(-1))
Gold (atomic radius = 0.150nm) crystallises in a face centred unit cell. What is the length of the side of the cell ?
Calculate the density of silver which crystallises in face-centred cubic from. The distance between nearest metal atoms is 287 pm (Molar mass of Ag = 107.87gmol^(-1),(N_(0)=6.022xx10^(23)mol^(-1)) .