Question

The number of atoms in 100 g of an fcc crystal with density $=10.0gc{m}^{-3}$ and cell edge equal to 200 pm is equal to:

• A. $5\times {10}^{24}$
• B. $5\times {10}^{25}$
• C. $6\times {10}^{23}$
• D. $2\times {10}^{25}$

Answer 1

The correct option is A $5\times {10}^{24}$

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

where $a$ is the edge of a unit cell

$N_A$ = Avogadro number $(6.022\times {10}^{23})$

$M$ = Molar mass

$Z$ = number of atoms per unit cell

For FCC,

$Z=4$

From the given data, we can calculate the molar mass of the given substance as:

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

​Number of atoms = $\frac{Givenmass}{molarmass\left(M\right)}\times N_A$

Putting the value of M in the above equation we get-

Number of atoms $=\frac{Givenmass\times Z\times N_A}{d\times a^3\times N_A}$

Number of atoms $=\frac{Givenmass\times Z}{d\times a^3}$

Putting the values we get:

Number of atoms$=\frac{100\times 4}{10\times (200\times {10}^{-10}{)}^3}$

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

Hence, option A is correct.