QUARTER FINAL STAGE 2026
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QUESTION
Find the density of an element of atomic mass $27$ atomic mass units, which crystallizes with a face-centered cubic structure of lattice constant $405$ picometers.
You may take one atomic mass unit to be equal to $1.66\times10^{-27}$ kg.
ANSWER: $2.7\times10^3$ kg/m³
SOLUTION 1:
Where:
$Z$ = Atoms per cell for FCC (4)
$M$ = Mass of one atom ($27 \times 1.66 \times 10^{-27}$ kg)
$a$ = Lattice constant ($405 \times 10^{-12}$ m)
Calculation:
$\rho = \dfrac{Z \cdot M}{a^3}$
$\rho = \dfrac{4 \times (4.482 \times 10^{-26})}{(4.05 \times 10^{-10})^3}$
$\rho = \dfrac{1.7928 \times 10^{-25}}{6.643 \times 10^{-29}}$
$\rho = 0.27 \times 10^4$
$\rho = 2.7 \times 10^3 \text{ kg/m³}$
SOLUTION 2:
Where:
$Z$ = Atoms per cell for FCC (4)
$M$ = Mass of one atom ($27 \times 1.7 \times 10^{-27}$ kg)
$a$ = Lattice constant ($4.1 \times 10^{-10}$ m)
Calculation:
$\rho = \dfrac{Z \cdot M}{a^3}$
$\rho = \dfrac{4 \times (45.9 \times 10^{-27})}{(4.1 \times 10^{-10})^3}$
$\rho = \dfrac{183.6 \times 10^{-27}}{68.9 \times 10^{-30}}$
$\rho = 2.7 \times 10^3 \text{ kg/m³}$
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PRACTICE QUESTIONS
1. QUESTION: Find the density of an element of atomic mass 128 atomic mass units, which crystallizes with a body-centred cubic structure of lattice constant 0.400 nm.
ANSWER: 6640 kg/m³ or $6.64\times10^{3}$ kg/m³
SOLUTION:
$Z = 2$ atoms per unit cell for a body-centred cubic structure
$m = 128 \times 1.66\times10^{-27}\text{ kg}$
$a = 4.00\times10^{-10}\text{ m}$
$a^3 = (4.00\times10^{-10})^3$
$a^3 = 6.40\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{2 \times 128 \times 1.66\times10^{-27}}{6.40\times10^{-29}}$
$\rho = \dfrac{256 \times 1.66\times10^{-27}}{64\times10^{-30}}$
$\rho = 4 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 4 \times 1.66\times10^{3}$
$\rho = 4 \times 166\times10^{1}$
$\rho = 6.64\times10^{3}\text{ kg/m}^3$
2. QUESTION: Find the density of an element of atomic mass 27.0 atomic mass units, which crystallizes with a simple cubic structure of lattice constant 0.300 nm.
ANSWER: 1660 kg/m³ or $1.66\times10^{3}$ kg/m³
SOLUTION:
$Z = 1$ atom per unit cell for a simple cubic structure
$m = 27.0 \times 1.66\times10^{-27}\text{ kg}$
$a = 3.00\times10^{-10}\text{ m}$
$a^3 = (3.00\times10^{-10})^3$
$a^3 = 2.70\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{27.0 \times 1.66\times10^{-27}}{2.70\times10^{-29}}$
$\rho = \dfrac{27 \times 1.66\times10^{-27}}{27\times10^{-30}}$
$\rho = 1.66\times10^{-27} \times10^{30}$
$\rho = 1.66\times10^{3}$
$\rho = 1.66\times10^{3}\text{ kg/m}^3$
3. QUESTION: Find the density of an element of atomic mass 108 atomic mass units, which crystallizes with a face-centred cubic structure of lattice constant 0.60 nm.
ANSWER: 3300 kg/m³ or $3.3\times10^{3}$ kg/m³
SOLUTION:
$Z = 4$ atoms per unit cell for a face-centred cubic structure
$m = 108 \times 1.66\times10^{-27}\text{ kg}$
$a = 6.0\times10^{-10}\text{ m}$
$a^3 = (6.0\times10^{-10})^3$
$a^3 = 2.16\times10^{-28}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{4 \times 108 \times 1.66\times10^{-27}}{2.16\times10^{-28}}$
$\rho = \dfrac{432 \times 1.66\times10^{-27}}{216\times10^{-30}}$
$\rho = 2 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 2 \times 1.66\times10^{3}$
$\rho = 2 \times 166\times10^{1}$
$\rho \approx 3.3\times10^{3}\text{ kg/m}^3$
4. QUESTION: Find the density of an element of atomic mass 24.0 atomic mass units, which crystallizes with a body-centred cubic structure of lattice constant 0.200 nm.
ANSWER: 9960 kg/m³ or $9.96\times10^{3}$ kg/m³
SOLUTION:
$Z = 2$ atoms per unit cell for a body-centred cubic structure
$m = 24.0 \times 1.66\times10^{-27}\text{ kg}$
$a = 2.00\times10^{-10}\text{ m}$
$a^3 = (2.00\times10^{-10})^3$
$a^3 = 8.00\times10^{-30}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{2 \times 24.0 \times 1.66\times10^{-27}}{8.00\times10^{-30}}$
$\rho = \dfrac{48 \times 1.66\times10^{-27}}{8\times10^{-30}}$
$\rho = 6 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 6 \times 1.66\times10^{3}$
$\rho = 6 \times 166\times10^{1}$
$\rho = 9.96\times10^{3}\text{ kg/m}^3$
5. QUESTION: Find the density of an element of atomic mass 162 atomic mass units, which crystallizes with a face-centred cubic structure of lattice constant 0.600 nm.
ANSWER: 4980 kg/m³ or $4.98\times10^{3}$ kg/m³
SOLUTION:
$Z = 4$ atoms per unit cell for a face-centred cubic structure
$m = 162 \times 1.66\times10^{-27}\text{ kg}$
$a = 6.00\times10^{-10}\text{ m}$
$a^3 = (6.00\times10^{-10})^3$
$a^3 = 2.16\times10^{-28}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{4 \times 162 \times 1.66\times10^{-27}}{2.16\times10^{-28}}$
$\rho = \dfrac{648 \times 1.66\times10^{-27}}{216\times10^{-30}}$
$\rho = 3 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 3 \times 1.66\times10^{3}$
$\rho = 3 \times 166\times10^{1}$
$\rho = 4.98\times10^{3}\text{ kg/m}^3$
6. QUESTION: Find the density of an element of atomic mass 189 atomic mass units, which crystallizes with a simple cubic structure of lattice constant 0.30 nm.
ANSWER: 12000 kg/m³ or $1.2\times10^{4}$ kg/m³
SOLUTION:
$Z = 1$ atom per unit cell for a simple cubic structure
$m = 189 \times 1.7\times10^{-27}\text{ kg}$
$a = 3.0\times10^{-10}\text{ m}$
$a^3 = (3.0\times10^{-10})^3$
$a^3 = 2.7\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{189 \times 1.7\times10^{-27}}{2.7\times10^{-29}}$
$\rho = \dfrac{189 \times 1.7\times10^{-27}}{27\times10^{-30}}$
$\rho = 7 \times 1.7\times10^{-27} \times10^{30}$
$\rho = 7 \times 1.7\times10^{3}$
$\rho = 7 \times 17\times10^{2}$
$\rho \approx 1.2\times10^{4}\text{ kg/m}^3$
7. QUESTION: Find the density of an element of atomic mass 32.0 atomic mass units, which crystallizes with a body-centred cubic structure of lattice constant 0.40 nm.
ANSWER: 1700 kg/m³ or $1.7\times10^{3}$ kg/m³
SOLUTION:
$Z = 2$ atoms per unit cell for a body-centred cubic structure
$m = 32.0 \times 1.7\times10^{-27}\text{ kg}$
$a = 4.0\times10^{-10}\text{ m}$
$a^3 = (4.0\times10^{-10})^3$
$a^3 = 6.4\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{2 \times 32.0 \times 1.7\times10^{-27}}{6.4\times10^{-29}}$
$\rho = \dfrac{64 \times 1.7\times10^{-27}}{64\times10^{-30}}$
$\rho = 1.7\times10^{-27} \times10^{30}$
$\rho = 1.7\times10^{3}$
$\rho = 1.7\times10^{3}\text{ kg/m}^3$
8. QUESTION: Find the density of an element of atomic mass 216 atomic mass units, which crystallizes with a face-centred cubic structure of lattice constant 0.60 nm.
ANSWER: 6600 kg/m³ or $6.6\times10^{3}$ kg/m³
SOLUTION:
$Z = 4$ atoms per unit cell for a face-centred cubic structure
$m = 216 \times 1.66\times10^{-27}\text{ kg}$
$a = 6.0\times10^{-10}\text{ m}$
$a^3 = (6.0\times10^{-10})^3$
$a^3 = 2.16\times10^{-28}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{4 \times 216 \times 1.66\times10^{-27}}{2.16\times10^{-28}}$
$\rho = \dfrac{864 \times 1.66\times10^{-27}}{216\times10^{-30}}$
$\rho = 4 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 4 \times 1.66\times10^{3}$
$\rho = 4 \times 166\times10^{1}$
$\rho \approx 6.6\times10^{3}\text{ kg/m}^3$
9. QUESTION: Find the density of an element of atomic mass 135 atomic mass units, which crystallizes with a simple cubic structure of lattice constant 0.30 nm.
ANSWER: 8300 kg/m³ or $8.3\times10^{3}$ kg/m³
SOLUTION:
$Z = 1$ atom per unit cell for a simple cubic structure
$m = 135 \times 1.66\times10^{-27}\text{ kg}$
$a = 3.0\times10^{-10}\text{ m}$
$a^3 = (3.0\times10^{-10})^3$
$a^3 = 2.7\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{135 \times 1.66\times10^{-27}}{2.7\times10^{-29}}$
$\rho = \dfrac{135 \times 1.66\times10^{-27}}{27\times10^{-30}}$
$\rho = 5 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 5 \times 1.66\times10^{3}$
$\rho = 5 \times 166\times10^{1}$
$\rho = 8.3\times10^{3}\text{ kg/m}^3$
10. QUESTION: Find the density of an element of atomic mass 54.0 atomic mass units, which crystallizes with a simple cubic structure of lattice constant 0.300 nm.
ANSWER: 3320 kg/m³ or $3.32\times10^{3}$ kg/m³
SOLUTION:
$Z = 1$ atom per unit cell for a simple cubic structure
$m = 54.0 \times 1.66\times10^{-27}\text{ kg}$
$a = 3.00\times10^{-10}\text{ m}$
$a^3 = (3.00\times10^{-10})^3$
$a^3 = 2.70\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{54.0 \times 1.66\times10^{-27}}{2.70\times10^{-29}}$
$\rho = \dfrac{54 \times 1.66\times10^{-27}}{27\times10^{-30}}$
$\rho = 2 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 2 \times 1.66\times10^{3}$
$\rho = 2 \times 166\times10^{1}$
$\rho = 3.32\times10^{3}\text{ kg/m}^3$
11. QUESTION: Find the density of an element of atomic mass 64.0 atomic mass units, which crystallizes with a body-centred cubic structure of lattice constant 0.40 nm.
ANSWER: 3300 kg/m³ or $3.3\times10^{3}$ kg/m³
SOLUTION:
$Z = 2$ atoms per unit cell for a body-centred cubic structure
$m = 64.0 \times 1.66\times10^{-27}\text{ kg}$
$a = 4.0\times10^{-10}\text{ m}$
$a^3 = (4.0\times10^{-10})^3$
$a^3 = 6.4\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{2 \times 64.0 \times 1.66\times10^{-27}}{6.4\times10^{-29}}$
$\rho = \dfrac{128 \times 1.66\times10^{-27}}{64\times10^{-30}}$
$\rho = 2 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 2 \times 1.66\times10^{3}$
$\rho = 2 \times 166\times10^{1}$
$\rho \approx 3.3\times10^{3}\text{ kg/m}^3$
12. QUESTION: Find the density of an element of atomic mass 96.0 atomic mass units, which crystallizes with a face-centred cubic structure of lattice constant 0.400 nm.
ANSWER: 9960 kg/m³ or $9.96\times10^{3}$ kg/m³
SOLUTION:
$Z = 4$ atoms per unit cell for a face-centred cubic structure
$m = 96.0 \times 1.66\times10^{-27}\text{ kg}$
$a = 4.00\times10^{-10}\text{ m}$
$a^3 = (4.00\times10^{-10})^3$
$a^3 = 6.40\times10^{-29}\text{ m}^3$
$\rho = \dfrac{Zm}{a^3}$
$\rho = \dfrac{4 \times 96.0 \times 1.66\times10^{-27}}{6.40\times10^{-29}}$
$\rho = \dfrac{384 \times 1.66\times10^{-27}}{64\times10^{-30}}$
$\rho = 6 \times 1.66\times10^{-27} \times10^{30}$
$\rho = 6 \times 1.66\times10^{3}$
$\rho = 6 \times 166\times10^{1}$
$\rho = 9.96\times10^{3}\text{ kg/m}^3$