Question

A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of $400\, pm$. If the density of the substance in the crystal is $8\, g \,cm^{- 3}$ , then the number of atoms present in $128\, g$ of the crystal is

• A. $2\times 10^{24}$
• B. $4\times 10^{24}$
• C. $1\times 10^{24}$
• D. $3\times 10^{24}$

Answer 1

Answer: (C)

$d=\frac{Z \times M}{a^{3} \times N_{A}}\Rightarrow M=\frac{a^{3} \times d \times N_{A}}{Z}$
$M=\frac{(400)^3\times 1 0^{- 30} \times 8 \times N_{A}}{4}$
On solving $M=64\times 2\times 10^{- 24}$ $N_{A}g$
Now, $\left(64 \times 2 \times 1 0^{- 24} N_{A}\right)$ g contains $N_{A}$ atoms
$128 \, g \,$ will contain $=\frac{N_{A}}{64 \times 2 \times 10^{- 24} N_{A}}\times 128$
$=1\times 10^{24}$