An element crystallizes in fcc lattice with cell edge of 300 pm. The density of element is 10.8gcm^{-3}. Calculate the number of atoms in 108g of the element.

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

Given:

FCC lattice, z=4

\rho = 10.8gcm^{-3}

a=300pm = 300\times 10^{-10}cm

\rho =\frac{z\times M}{a^{3}\times N_{A}}

M = \frac{\rho \times a^{3\times N_{A}}}{z}

        = \frac{10.8\times \left ( 300\times 10^{-10} \right )\times 6.022\times 10^{23}}{4}=175.6g

n = \frac{108}{175.6}=0.615mol

1 mol contains → N_{A} atoms

0.615 mol → 0.615\times N_{A} = 14.81 \times 10^{23}atoms

 

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