12. Element X crystallizes into an FCC lattice with cell edge of 200 pm. The
density of the element is 5 g/cm3. 200 g of this element contain how many atoms?
a. 2.0 x 1023
b. 2.0 x 1025
c. 6.02 x 1023
d. 6.02 x 1025
Element X in an FCC lattice with a 200 pm cell edge and 5 g/cm³ density yields approximately 2.0 × 1025 atoms in 200 g, matching option b. This calculation uses the FCC unit cell’s 4 atoms, unit cell volume, and density to find total atoms precisely.
Step-by-Step Solution
- Convert cell edge to cm: 200 pm = 200 × 10-10 cm = 2 × 10-8 cm
- Unit cell volume V = a³ = (2 × 10-8)³ = 8 × 10-24 cm³
- Mass per unit cell = density × V = 5 × 8 × 10-24 = 4 × 10-23 g
- Number of unit cells in 200 g = 200 / (4 × 10-23) = 5 × 1024
- Total atoms = unit cells × 4 (FCC atoms) = 5 × 1024 × 4 = 2.0 × 1025
Option Analysis
| Option | Value | Why Incorrect/Correct? |
|---|---|---|
| a. 2.0 × 1023 | 2.0 × 1023 | Matches 200 g of ~10 g/mol element (ignores density/lattice specifics) |
| b. 2.0 × 1025 | 2.0 × 1025 | Correct; matches calculated atoms |
| c. 6.02 × 1023 | 6.02 × 1023 | Avogadro’s number (1 mole); 200 g implies ~33 g/mol but mismatches FCC/density |
| d. 6.02 × 1025 | 6.02 × 1025 | ~10 moles; overestimates by using NA directly without unit cell adjustment |
SEO Article: FCC Lattice Density Calculation
In FCC lattice density calculations for exams like CSIR NET, determining atoms in a mass requires unit cell volume, density, and lattice type. This FCC lattice 200 pm edge 5 g/cm³ problem tests these fundamentals: Element X forms FCC with a = 200 pm, density = 5 g/cm³—find atoms in 200 g.
Core Formula
Atoms N = (total mass / unit cell mass) × Z, where Z = 4 for FCC[execute_python:1]. Unit cell mass = density × a³ (a in cm).
Common Pitfalls
- Forgetting pm-to-cm conversion leads to ~106 error.
- Using Z = 1 (simple cubic) halves result.
- Assuming 1 mole ignores derived atomic mass ~100 g/mol here.
Master FCC lattice density calculation for competitive exams—practice yields precision.
