Question

The number of atoms in $100$ g of fcc crystal with density $d=10gc{m}^{-3}$ and edge length of $200$ pm is:

• A. $3\times {10}^{25}$
• B. $5\times {10}^{24}$
• C. $1\times {10}^{25}$
• D. $2\times {10}^{25}$

Answer 1

Answer: (B)

$5\times {10}^{24}$

$Z_{effective}=4$ in FCC structure.

$Z_A=4;{\rho }_A=10gm/c{m}^3;a=200pm=200\times {10}^{-10}cm$

Molar mass, $M_A=?$

${\rho }_A=\frac{Z_A\times M_A}{N_A\times a^3}$

where $N_A$ is Avogadro number.

$M_A=\frac{10gc{m}^{-3}\times 6.02\times {10}^{23}\times {200}^3\times {10}^{-30}c{m}^3}{4}$

$M_A=12.046gm$

In one unit cell, $4$ atoms of $A$ are present.

Number of atoms in $100gm$ FCC $=no.ofmoles\times N_A=\frac{100}{12.046}\times 6.02\times {10}^{23}=5\times {10}^{24}atoms$