Question

The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is:

• A. $6.6×10^{20}$
• B. $6.6×10^{23}$
• C. $6.6×10^{19}$
• D. $6.6×10^{22}$

Hint:

The formula d= Z *M/(NA*a3) relates density to atomic mass and unit cell parameters.

Answer 1

The Correct Option is D

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

\( \implies M = \frac{10 \times 6.022 \times 10^{23} \times (300 \times 10^{-10})^3}{4} \)

\( M = 40.5 \, \text{gm} \)

Therefore, \( 40.5 \, \text{gm} \rightarrow 6.022 \times 10^{23} \, \text{atoms} \)

\( 4.5 \, \text{gm} \rightarrow x \)

\( x = 6.6 \times 10^{22} \, \text{atoms} \)