Question

The number of unit cells present in 5.85 g of a cube-shaped crystal of sodium chloride is \( x \times 10^y \). \( x \) and \( y \) respectively are (Na = 23 u, Cl = 35.5 u) \[ (N_A = 6 \times 10^{23} { mol}^{-1}) \]

• A. 15, 21
• B. 15, 22
• C. 30, 23
• D. 15, 23

Hint:

In calculations involving crystal structures, knowing the type of lattice (e.g., FCC, BCC) and the number of formula units per unit cell is crucial for determining the number of unit cells.

Answer 1

The Correct Option is A

Step 1: Calculate the number of moles of NaCl. Molar mass of NaCl: \[ 23 + 35.5 = 58.5 { g/mol} \] Moles of NaCl in 5.85 g: \[ \frac{5.85}{58.5} = 0.1 { moles} \] Number of NaCl formula units: \[ 0.1 \times (6 \times 10^{23}) = 6 \times 10^{22} \] Step 2: Determine the number of unit cells. In a face-centered cubic (FCC) unit cell, 4 NaCl formula units are present per unit cell. Number of unit cells: \[ \frac{6 \times 10^{22}}{4} = 1.5 \times 10^{21} \] Thus, \( x = 15 \) and \( y = 21 \), so the correct answer is \( 15, 21 \).