Question

Calculate the volume of unit cell if an element having molar mass $56 \,\text{g mol}^{-1}$ forms bcc unit cells. $[\rho \cdot N_A = 4.8 \times 10^{24} \,\text{g cm}^{-3}\text{mol}^{-1}]$

• A. $1.17\times10^{-23}cm^{3}$
• B. $4.79\times10^{-23}cm^{3}$
• C. $3.31\times10^{-23}cm^{3}$
• D. $2.33\times10^{-23}cm^{3}$

Hint:

Chemistry Tip: Always memorize the $Z$ values for common unit cells: Simple Cubic (sc) $Z=1$, Body-Centered Cubic (bcc) $Z=2$, and Face-Centered Cubic (fcc) $Z=4$.

Answer 1

The Correct Option is D

Concept: Chemistry (Solid State) - Density of Unit Cell.

Step 1: Identify the given parameters. The molar mass of the element is given as $M = 56~g~mol^{-1}$. The product of density and Avogadro's number is given as $\rho\cdot N_{A} = 4.8\times10^{24}~g~cm^{-3}mol^{-1}$.

Step 2: Determine the number of atoms per unit cell ($Z$). The element crystallizes in a body-centered cubic (bcc) structure. For a bcc unit cell, the total number of atoms per unit cell is $Z = 2$ (1 atom from the 8 corners + 1 atom fully at the body center).

Step 3: State the formula relating density, mass, and volume. The density ($\rho$) of a unit cell is calculated using the formula $\rho = \frac{Z \cdot M}{N_{A} \cdot V}$, where $V$ is the volume of the unit cell, and $N_{A}$ is Avogadro's number.

Step 4: Rearrange the formula to solve for volume ($V$). We need to find the volume, so we rearrange the equation: $V = \frac{Z \cdot M}{\rho \cdot N_{A}}$.

Step 5: Substitute the values and calculate the final result. Substitute the known values into the rearranged formula: $V = \frac{2 \times 56}{4.8 \times 10^{24}}$. First, calculate the numerator: $2 \times 56 = 112$.
Now, divide this by the denominator: $V = \frac{112}{4.8} \times 10^{-24}$. Performing the division $112 \div 4.8$ yields exactly $23.333...$. Thus, the volume is $23.33 \times 10^{-24} cm^3$.
To match the standard scientific notation format provided in the options, we shift the decimal point one place to the left, which increases the exponent by 1, resulting in the final value of $2.33 \times 10^{-23} cm^3$. $$ \therefore \text{The volume of the unit cell is } 2.33\times10^{-23}cm^{3}. $$