Question

Calculate the number of atoms present in 90 g of metal if it forms bcc structure.
[ \(ρ\times a^3 = 1.8\times 10^{-22}\) g ]

• A. \(4.0\times 10^{24}\)
• B. \(1.0\times 10^{24}\)
• C. \(2.0\times 10^{24}\)
• D. \(3.0\times 10^{24}\)

Hint:

Use density formula to get molar mass, then moles and atoms.

Answer 1

The Correct Option is B

Step 1: Mass of one atom:
A bcc cell holds 2 atoms and has mass $\rho a^3 = 1.8\times10^{-22}$ g. So one atom has mass $0.9\times10^{-22}$ g.

Step 2: Count atoms:
$N = \dfrac{90}{0.9\times10^{-22}} = 100\times10^{22} = 1.0\times10^{24}$.

Step 3: Result:
This is option (B). Avogadro's number is not even needed in this route.

Final Answer:
There are 1.0e24 atoms in 90 g. \[ \boxed{\text{(B) }1.0\times10^{24}} \]