Question

Calculate the density of metal having molar mass $210 \text{ g mol}^{-1}$ that forms simple cubic unit cell. (Given: $a^{3} \cdot N_{A} = 21.5 \text{ cm}^{3} \text{ mol}^{-1}$)}

• A. 9.77 \text{ g cm}^{-3}
• B. 7.15 \text{ g cm}^{-3}
• C. 8.12 \text{ g cm}^{-3}
• D. 6.94 \text{ g cm}^{-3}

Hint:

Always remember the formula for the density of a unit cell: $\rho = \frac{Z \cdot M}{a^3 \cdot N_A}$. Pay attention to the type of unit cell (simple cubic, BCC, FCC) to correctly determine the value of $Z$. Simple cubic: $Z=1$, Body-centered cubic (BCC): $Z=2$, Face-centered cubic (FCC): $Z=4$.

Answer 1

The Correct Option is C

Step 1: Recall the formula for the density ($\rho$) of a substance forming a cubic unit cell. The formula is: \[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] where:

• $Z$ is the number of atoms per unit cell.
• $M$ is the molar mass of the metal.
• $a$ is the edge length of the unit cell.
• $N_A$ is Avogadro's Number ($6.022 \times 10^{23} \text{ mol}^{-1}$).

Step 2: Identify the given values and unit cell type.

• The metal forms a simple cubic unit cell, so the number of atoms per unit cell, $Z = 1$.
• The molar mass ($M$) of the metal is $210 \text{ g mol}^{-1}$.
• We are given the product $a^{3} \cdot N_{A} = 21.5 \text{ cm}^{3} \text{ mol}^{-1}$.

Step 3: Substitute the given values into the density formula. \[ \rho = \frac{1 \cdot (210 \text{ g mol}^{-1})}{21.5 \text{ cm}^{3} \text{ mol}^{-1 \]
Step 4: Calculate the density. \[ \rho = \frac{210}{21.5} \text{ g cm}^{-3} \] \[ \rho \approx 9.7674 \text{ g cm}^{-3} \] Rounding to two decimal places, the density is $9.77 \text{ g cm}^{-3}$.