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

To solve the problem of calculating the number of atoms in a given mass of a face-centred cubic (FCC) crystal, we need to follow a series of steps involving understanding the crystal structure and performing some calculations.

1. Identify the unit cell type: The crystal is face-centred cubic (FCC). 
2. Determine the number of atoms per unit cell in FCC:
3. Understand the relation between edge length and atomic radius for FCC:
4. Calculate the volume of 4.5 g of the crystal:
5. Convert mass to moles:
6. Calculate the total number of atoms:
7. Conclusion:

To conclude, the problem involves understanding FCC structure, using geometric relations in cubic cells, converting given mass into mole with respect to atomic weight, and finally applying Avogadro's number for the total atom count.

Answer 2

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

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

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

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

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

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