Question

Calculate the number of atoms present in \(1 \text{g}\) metal that forms simple unit cell structure if product of density and volume of unit cell is \(66.5\times 10^{-24} \text{g}\)

• A. \(1.504\times 10^{22}\)
• B. \(3.419\times 10^{22}\)
• C. \(2.516\times 10^{22}\)
• D. \(2.018\times 10^{22}\)

Hint:

Density times volume is the mass of the unit cell, and a simple cubic cell holds one atom.

Answer 1

The Correct Option is A

Step 1: Link to molar mass:
$\rho a^3 = \frac{ZM}{N_A}$, with $Z = 1$, so the molar mass is $M = 66.5 \times 10^{-24} \times 6.022 \times 10^{23} = 40.05$ g/mol.

Step 2: Count atoms:
Moles in 1 g $= \frac{1}{40.05} = 0.02497$.
Atoms $= 0.02497 \times 6.022 \times 10^{23} = 1.504 \times 10^{22}$.

Final Answer:
Atoms in 1 g $= 1.504 \times 10^{22}$, option (A). \[ \boxed{1.504 \times 10^{22}} \]