The distance between `Na^+` and `Cl^-` ions in NaCl with a density `3.165 g cm^(-3)` is

Options

A. 497 pm
B. 248.5 pm
C. 234 pm
D. 538.5 pm

Explanation

To find the distance between Na⁺ and Cl⁻ ions in NaCl with a density of 3.165 g/cm³, we can follow these steps:

### Step 1: Understand the formula for density
The density (ρ) of a crystal structure can be expressed using the formula:
\[ \rho = \frac{Z \cdot m}{N_a \cdot A^3} \]
where:
- \( Z \) = number of formula units per unit cell
- \( m \) = molar mass of the compound
- \( N_a \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol⁻¹)
- \( A \) = edge length of the unit cell

### Step 2: Identify the values
For NaCl:
- The number of formula units per unit cell, \( Z = 4 \) (since NaCl has a face-centered cubic structure).
- The molar mass of NaCl = mass of Na + mass of Cl = 23 g/mol + 35.5 g/mol = 58.5 g/mol.
- The density \( \rho = 3.165 \, \text{g/cm}^3 \).

### Step 3: Rearrange the density formula to find \( A^3 \)
Rearranging the formula gives us:
\[ A^3 = \frac{Z \cdot m}{N_a \cdot \rho} \]

### Step 4: Substitute the values into the formula
Substituting the known values:
\[ A^3 = \frac{4 \cdot 58.5 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 3.165 \, \text{g/cm}^3} \]

Calculating the numerator:
\[ 4 \cdot 58.5 = 234 \, \text{g/mol} \]

Now substituting into the equation:
\[ A^3 = \frac{234}{6.022 \times 10^{23} \cdot 3.165} \]

Calculating the denominator:
\[ 6.022 \times 10^{23} \cdot 3.165 \approx 1.907 \times 10^{24} \]

Now, substituting back:
\[ A^3 = \frac{234}{1.907 \times 10^{24}} \approx 1.227 \times 10^{-22} \, \text{cm}^3 \]

### Step 5: Calculate \( A \)
Taking the cube root to find \( A \):
\[ A \approx (1.227 \times 10^{-22})^{1/3} \approx 4.97 \times 10^{-8} \, \text{cm} \]

### Step 6: Convert \( A \) to picometers
To convert centimeters to picometers:
\[ A \approx 4.97 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 497 \, \text{pm} \]

### Step 7: Calculate the distance between Na⁺ and Cl⁻ ions
In NaCl, the distance between Na⁺ and Cl⁻ ions is half of the edge length \( A \):
\[ \text{Distance} = \frac{A}{2} = \frac{497 \, \text{pm}}{2} = 248.5 \, \text{pm} \]

### Final Answer
The distance between Na⁺ and Cl⁻ ions in NaCl is approximately **248.5 pm**.

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