Question

An element crystallises in fcc unit cell with cell edge length of \(3.608\times 10^{-8}\) cm, the density of element is \(8.92\text{ gcm}^{-3}\). Calculate the atomic mass of element (\(\text{N}_A = 6.022\times 10^{23}\)).

• A. \(60\) g/mol
• B. \(65\) g/mol
• C. \(63\) g/mol
• D. \(108\) g/mol

Hint:

Use M = rho a^3 N_A / Z with Z = 4 for fcc.

Answer 1

The Correct Option is C

Step 1: Mass in one cell:
Mass of one unit cell $=\rho a^3 = 8.92\times 4.697\times 10^{-23} = 4.190\times 10^{-22}$ g.

Step 2: Atoms in one cell:
An fcc cell has 8 corners at 1/8 each and 6 faces at 1/2 each, so $1 + 3 = 4$ atoms.

Step 3: Mass of one atom:
$4.190\times 10^{-22}/4 = 1.0475\times 10^{-22}$ g.

Step 4: Mass of one mole:
Multiply by Avogadro number: $1.0475\times 10^{-22}\times 6.022\times 10^{23} = 63.08$ g.

Step 5: Match:
This is option (C), 63 g/mol.

Final Answer:
One mole of atoms weighs about 63 g. \[ \boxed{C} \]