Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Silver crystallises in fcc lattice.If edge length of the cell is $4.077\times 10^{-8}cm$ and density is $10.5gcm^{-3}$.Calculate the atomic mass of silver.

$\begin{array}{1 1}107.08gmol^{-1}\\207.08gmol^{-1}\\307.08gmol^{-1}\\407.08gmol^{-1}\end{array} $

Can you answer this question?

1 Answer

0 votes
$M=\large\frac{d\times a^3\times N_A}{z}$
Where $d$=Density of the material
$a$=Length of the edge of the cell
$N_A$=Avagadro number
$z$=No. of atoms
$M=\large\frac{10.5gcm^{-3}\times (4.077\times 10^{-8}cm)^3\times (6.023\times 10^{23}mol^{-1})}{4}$
$\therefore$ Atomic mass of silver =$107.08 gmol^{-1}$
answered Jul 30, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App