Question

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^-3. The molar mass of the metal is
(N_A Avogadro's constant = 6.02 × 10²³ mol^-1)

• A. 40 g mol^-1
• B. 30 g mol^-1
• C. 27 g mol^-1
• D. 20 g mol^-1

Answer 1

The Correct Option is C

To determine the molar mass of the metal with a face-centered cubic (fcc) lattice, we can use the following formula relating density (\( \rho \)), the molar mass (\( M \)), Avogadro's number (\( N_A \)), and the edge length of the unit cell (\( a \)):

M = \frac{{\rho \cdot a^3 \cdot N_A}}{{Z}}

Where:

• \rho = 2.72 \, \text{g cm}^{-3}
• The edge length \( a = 404 \, \text{pm} = 404 \times 10^{-10} \, \text{cm} \) (since 1 pm = \(10^{-12}\) m = \(10^{-10}\) cm)
• \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
• For fcc, the number of atoms per unit cell \( Z = 4 \)

First, convert the unit cell volume \( a^3 \) into \( \text{cm}^3 \):

a^3 = (404 \times 10^{-10})^3 \, \text{cm}^3\

= 404^3 \times 10^{-30} \, \text{cm}^3\

= 65.5744 \times 10^{-24} \, \text{cm}^3\

Now, substitute the values into the formula to find the molar mass \( M \):

M = \frac{{2.72 \times 65.5744 \times 10^{-24} \times 6.02 \times 10^{23}}}{4}

Calculate the numerator:

= 2.72 \times 65.5744 \times 6.02 \times 10^{-1}

= 1071.05954528\

Then, calculate the molar mass:

M = \frac{107.105954528}{4} \approx 27 \, \text{g mol}^{-1}

Thus, the molar mass of the metal is 27 g mol^-1, corresponding to the correct option.