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: \[ \frac{6 \times 10^{22}}{4} = 15 \times 10^{21} \] Thus, \( x = 15 \) and \( y = 21 \), so the correct answer is \( 15, 21 \).

Answer 2

To solve the problem, we need to determine the number of unit cells present in 5.85 g of a cube-shaped crystal of sodium chloride (NaCl).

1. Given Data:
We are given:

• Mass of NaCl = 5.85 g
• Molar mass of NaCl = 58.5 g/mol
• Avogadro's number = \( 6 \times 10^{23} \, \text{mol}^{-1} \)
• The unit cell type of NaCl is assumed to be face-centered cubic (FCC), where each unit cell contains 4 formula units.

2. Number of Moles of NaCl:
To calculate the number of moles of NaCl, we use the formula: \[ \text{Moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar Mass of NaCl}} = \frac{5.85}{58.5} = 0.1 \, \text{mol} \]

3. Number of Formula Units in 0.1 Moles:
The number of formula units in 0.1 moles of NaCl is given by: \[ \text{Number of formula units} = \text{Moles of NaCl} \times \text{Avogadro's number} = 0.1 \times 6 \times 10^{23} = 6 \times 10^{22} \]

4. Number of Unit Cells:
Since each FCC unit cell of NaCl contains 4 formula units, the number of unit cells is: \[ \text{Number of unit cells} = \frac{\text{Number of formula units}}{4} = \frac{6 \times 10^{22}}{4} = 1.5 \times 10^{22} \]

5. Final Answer:
The number of unit cells present in 5.85 g of NaCl is \( 15 \times 10^{21} \), which corresponds to the option (1).

Final Answer:
The correct option is (A) 15, 21.