Question

Cells present in a cube-shaped ideal crystal of NaCl of mass 1.00 g ? [Atomic masses : Na = 23, Cl = 35.5]

• A. $5.14 \times 10^{21}$ unit cells
• B. $1.28 \times 10^{21}$ unit cells
• C. $1.71 \times 10^{21}$ unit cells
• D. $2.57 \times 10^{21}$ unit cells

Answer 1

Answer: (D)

$2.57 \times 10^{21}$ unit cells

Answer 2

Mass of one unit-cell (m) = volume $\times$ density $=a^{3} \times d$ $=a^{3} \times \frac{M z}{N_{0} a^{3}}=\frac{M z}{N_{0}}$ $m =\frac{58.5 \times 4}{6.02 \times 10^{23}} g$ $\therefore$ Number of unit cells in $1 g =\frac{1}{m}$ $=\frac{6.02 \times 10^{23}}{58.5 \times 4}$ $=2.57 \times 10^{21}$