The number of atoms in 100 g of an FCC crystal with density d = 10 g cm−3 and cell edge as 200 pm is equal to:

The number of atoms in 100 g of an FCC crystal with density d = 10 g ${cm}^{- 3}$ and cell edge as 200 pm is equal to:

A
$3 \times 10^{25}$
B
$0.5 \times 10^{25}$
C
$1 \times 10^{25}$
D
$2 \times 10^{25}$

Solution:

$M = \frac{ρ \times a^{3} \times N_{A} \times 10^{- 30}}{n}$

$= \frac{10 \times {(200)}^{3} \times (6.02 \times 10^{23}) \times 10^{- 30}}{4}$

= 12.04

No. of atoms is 100 g = $\frac{6.02 \times 10^{23}}{12.04} \times 100$

$= 0.5 \times 10^{25}$

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