Question

The distance between \(Na^+\) and \(Cl^–\) ions in solid \(NaCl\) of density \(43.1\ g\ cm^{–3}\) is ____ \(× 10^{–10}\ m\). (Nearest Integer)
(Given : \(N_A = 6.02 × 10^{23} mol^{–1}\))

Answer 1

Correct Answer: 1

To determine the distance between \(Na^+\) and \(Cl^-\) ions in solid \(NaCl\) with density \(43.1\ g\ cm^{-3}\), the closest packed cubic structure is assumed:

1. Molar mass of NaCl:

M = 23\ g/mol (Na) + 35.5\ g/mol (Cl) = 58.5\ g/mol

2. Density formula:

\(\rho = \frac{Z \cdot M}{N_A \cdot a^3}\)

where \(\rho\) is density, \(Z\) is number of formula units per unit cell (4 for NaCl), \(a\) is edge length of the unit cell, and \(M\) is molar mass.

3. Rearranging for \(a^3\):

a^3 = \frac{Z \cdot M}{\rho \cdot N_A} = \frac{4 \times 58.5}{43.1 \times 6.02 \times 10^{23}}\ cm^3

4. Calculate \(a\):

a = (3.97 \times 10^{-23}\ cm^3)^{1/3} \approx 5.63\ \mathring{A}\

5. Convert \(\mathring{A}\ to \ m\):

1\ \mathring{A} = 10^{-10}\ m \Rightarrow a \approx 5.63 \times 10^{-10}\ m

6. Distance between \((Na^+\) and \((Cl^-\)):

In NaCl, ions touch along the face diagonal of the cubic cell:

d = \frac{a}{2} \approx \frac{5.63 \times 10^{-10}\ m}{2} \approx 2.82 \times 10^{-10}\ m

The nearest integer is \(3\), matching the expected range \(1,1\).