Question

Arrange the following elements with different atomic weight and same density according to decreasing number of atoms per unit volume:
A. Atomic weight \(=10\),
B. Atomic weight \(=20\),
C. Atomic weight \(=30\),
D. Atomic weight \(=40\).

• A. A \(>\) B \(>\) D \(>\) C
• B. A \(>\) B \(>\) C \(>\) D
• C. D \(>\) C \(>\) A \(>\) B
• D. D \(>\) C \(>\) B \(>\) A

Hint:

For equal density, number of atoms per unit volume varies inversely with atomic weight.

Answer 1

The Correct Option is B

Concept:
Number of atoms per unit volume is inversely proportional to atomic weight when density is the same. \[ n=\frac{\rho N_A}{M} \] where, \[ \rho=\text{density} \] \[ N_A=\text{Avogadro's number} \] \[ M=\text{atomic weight} \]

Step 1: Since density is same.
\[ n\propto \frac{1}{M} \]

Step 2: Compare atomic weights. \[ A=10,\quad B=20,\quad C=30,\quad D=40 \] Smaller atomic weight gives larger number of atoms per unit volume.

Step 3: Arrange in decreasing order. \[ 10<20<30<40 \] So number of atoms per unit volume is: \[ A>B>C>D \] \[ \therefore \text{Correct Answer is (B)} \]