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

To begin with, we need to find the distance between $$Na^+$$ and $$Cl^-$$ ions in solid NaCl. Recall that NaCl crystallizes in a face-centered cubic (FCC) structure and that each unit cell contains four formula units of NaCl, so the number of formula units per unit cell is $$Z = 4$$.

Next, the density of a crystal is given by
$$\rho = \frac{Z \times M}{N_A \times a^3}$$
where $$Z = 4$$, $$M = 23 + 35.5 = 58.5$$ g mol$$^{-1}$$ is the molar mass of NaCl, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, and $$\rho = 43.1$$ g cm$$^{-3}$$. Rearranging for $$a^3$$ yields
$$a^3 = \frac{Z \times M}{N_A \times \rho} = \frac{4 \times 58.5}{6.02 \times 10^{23} \times 43.1}$$ which in turn gives
$$a^3 = \frac{234}{2.5946 \times 10^{25}} = 9.02 \times 10^{-24} \text{ cm}^3$$ and hence
$$a = (9.02 \times 10^{-24})^{1/3} = 2.08 \times 10^{-8} \text{ cm}.$$

Then, in the NaCl FCC structure, the $$Na^+$$ and $$Cl^-$$ ions alternate along each edge of the unit cell, so the nearest-neighbor distance (distance between adjacent $$Na^+$$ and $$Cl^-$$) is exactly half the edge length of the unit cell:
$$d = \frac{a}{2} = \frac{2.08 \times 10^{-8} \text{ cm}}{2} = 1.04 \times 10^{-8} \text{ cm}.$$

Finally, since $$1 \text{ cm} = 10^{-2} \text{ m}$$, it follows that
$$d = 1.04 \times 10^{-8} \text{ cm} = 1.04 \times 10^{-10} \text{ m}.$$ Expressing this result as $$d = n \times 10^{-10}$$ m gives $$n \approx 1$$ (to the nearest integer).

The correct answer is $$\mathbf{1}$$.