The distance between `Na^(+) and Cl^(-) ` ions in solid NaCl of density `43.1 " g "cm^(-3)` is `__________ xx 10^(-10)` m.
`( "Given " N_(A) = 6.02 xx 10^(23) "mol"^(-1))`

Explanation

To find the distance between Na⁺ and Cl⁻ ions in solid NaCl, we can use the formula for density and the properties of the crystal structure of NaCl. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Understand the Crystal Structure:**
- NaCl crystallizes in a face-centered cubic (FCC) structure. In this structure, Cl⁻ ions are located at the corners and the centers of the faces of the cube, while Na⁺ ions occupy the octahedral voids.

2. **Use the Density Formula:**
- The density (d) of a substance can be expressed as:
\[ d = \frac{Z \cdot M}{N_A \cdot V} \]
- Where:
- \( Z \) = number of formula units per unit cell (for NaCl, \( Z = 4 \))
- \( M \) = molar mass of NaCl (approximately 58.5 g/mol)
- \( N_A \) = Avogadro's number (\( 6.02 \times 10^{23} \, \text{mol}^{-1} \))
- \( V \) = volume of the unit cell (which is \( a^3 \), where \( a \) is the edge length of the cube)

3. **Rearranging the Density Formula:**
- We can rearrange the formula to find the edge length \( a \):
\[ V = \frac{Z \cdot M}{N_A \cdot d} \]
\[ a^3 = \frac{Z \cdot M}{N_A \cdot d} \]
\[ a = \left( \frac{Z \cdot M}{N_A \cdot d} \right)^{1/3} \]

4. **Substituting the Values:**
- Given:
- \( d = 43.1 \, \text{g/cm}^3 \)
- \( M = 58.5 \, \text{g/mol} \)
- \( Z = 4 \)
- \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- Substitute these values into the equation:
\[ a = \left( \frac{4 \cdot 58.5}{6.02 \times 10^{23} \cdot 43.1} \right)^{1/3} \]

5. **Calculating the Edge Length (a):**
- Calculate the numerator:
\[ 4 \cdot 58.5 = 234 \]
- Calculate the denominator:
\[ 6.02 \times 10^{23} \cdot 43.1 \approx 2.594 \times 10^{25} \]
- Now calculate \( a^3 \):
\[ a^3 = \frac{234}{2.594 \times 10^{25}} \approx 9.02 \times 10^{-24} \, \text{cm}^3 \]
- Taking the cube root:
\[ a \approx (9.02 \times 10^{-24})^{1/3} \approx 2.08 \times 10^{-8} \, \text{cm} \]

6. **Finding the Distance Between Ions:**
- The distance between Na⁺ and Cl⁻ ions is half the edge length:
\[ \text{Distance} = \frac{a}{2} \approx \frac{2.08 \times 10^{-8}}{2} \approx 1.04 \times 10^{-8} \, \text{cm} \]
- Convert to meters:
\[ 1.04 \times 10^{-8} \, \text{cm} = 1.04 \times 10^{-10} \, \text{m} \]

### Final Answer:
The distance between Na⁺ and Cl⁻ ions in solid NaCl is approximately \( 1.04 \times 10^{-10} \, \text{m} \).

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