Question

The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is :
(Given density=10gcm^-3 and N_A=6.022×10²³).

• A. 6.6×10²⁰
• B. 6.6×10²³
• C. 6.6×10¹⁹
• D. 6.6×10²²

Answer 1

The Correct Option is D

We are given:
Mass of the crystal = 4.5 g
Density of the crystal \( \rho = 10 \, \text{g/cm}^3 \)
Avogadro's number \( N_A = 6.022 \times 10^{23} \)
Edge length of the FCC crystal \( a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m} \)
Step 1: Calculate the volume of the crystal The volume of the crystal can be calculated from the density and mass using the formula: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{4.5 \, \text{g}}{10 \, \text{g/cm}^3} = 0.45 \, \text{cm}^3 \] Step 2: Calculate the number of unit cells in the crystal For a face-centred cubic (FCC) crystal, the number of atoms per unit cell is 4. To find the number of unit cells in the crystal, we first calculate the volume of one unit cell. The volume of the unit cell for an FCC crystal is related to the edge length \( a \) by the formula: \[ V_{\text{unit cell}} = a^3 \] where \( a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m} \). Convert the edge length into cm: \[ a = 300 \times 10^{-12} \, \text{m} = 300 \times 10^{-10} \, \text{cm} \] Now, calculate the volume of one unit cell: \[ V_{\text{unit cell}} = (300 \times 10^{-10})^3 = 27 \times 10^{-30} \, \text{cm}^3 \] Step 3: Calculate the number of unit cells in the crystal Now, divide the total volume of the crystal by the volume of one unit cell to find the number of unit cells: \[ \text{Number of unit cells} = \frac{\text{Volume of crystal}}{\text{Volume of one unit cell}} = \frac{0.45 \, \text{cm}^3}{27 \times 10^{-30} \, \text{cm}^3} = 1.67 \times 10^{28} \, \text{unit cells} \] Step 4: Calculate the total number of atoms Since each unit cell contains 4 atoms (FCC structure), the total number of atoms in the crystal is: \[ \text{Total number of atoms} = 4 \times 1.67 \times 10^{28} = 6.68 \times 10^{28} \, \text{atoms} \] Thus, the number of atoms in 4.5 g of the FCC crystal is approximately \( 6.6 \times 10^{22} \) atoms.

Therefore, the correct answer is (D) \( 6.6 \times 10^{22} \).