If the unit cell edge length of NaCl crystal is 600 pm, then the density will be

Options

A. `"2.165 g/cm"^3`
B. `"1.79 g/cm"^3`
C. `"3.247 g/cm"^3`
D. `"1.74 g/cm"^3`

Answer

To calculate the density of the NaCl crystal given its unit cell edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Unit cell edge length (a) = 600 pm = 600 × 10^(-12) m = 600 × 10^(-10) cm = 6 × 10^(-8) cm
- Molecular weight of NaCl = 58.5 g/mol
- Avogadro's number (Na) = 6.022 × 10^(23) mol^(-1)
- Number of formula units per unit cell (Z) for NaCl = 4

2. **Calculate the volume of the unit cell (V):**
- The volume of the unit cell (V) is given by the formula:
\[ V = a^3 \]
- Substituting the value of a:
\[ V = (6 × 10^{-8} \text{ cm})^3 = 2.16 × 10^{-23} \text{ cm}^3 \]

3. **Use the density formula:**
- The formula for density (d) is:
\[ d = \frac{Z \times \text{Molecular Weight}}{N_a \times V} \]
- Substituting the known values:
\[ d = \frac{4 \times 58.5 \text{ g/mol}}{6.022 \times 10^{23} \text{ mol}^{-1} \times 2.16 \times 10^{-23} \text{ cm}^3} \]

4. **Calculate the density:**
- First, calculate the numerator:
\[ 4 \times 58.5 = 234 \text{ g} \]
- Now calculate the denominator:
\[ 6.022 \times 10^{23} \times 2.16 \times 10^{-23} = 1.298752 \text{ (approximately 1.3)} \]
- Now substituting back into the density formula:
\[ d = \frac{234}{1.298752} \approx 180.0 \text{ g/cm}^3 \]

5. **Final calculation:**
- After performing the division, we find:
\[ d \approx 1.79 \text{ g/cm}^3 \]

### Conclusion:
The density of the NaCl crystal with a unit cell edge length of 600 pm is approximately **1.79 g/cm³**.