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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} $

1 Answer

$M=\large\frac{d\times a^3\times N_A}{z}$
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

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