An ionic compound MX crystallizes in a rock-salt lattice. The molar...

For a rock-salt (NaCl type) lattice, the effective number of formula units per unit cell is $Z = 4$.

Using the density formula:
$d = \dfrac{Z \times M}{N_A \times a^3}$

Substitute the given values to find the volume of the unit cell, $a^3$:
$5.0 = \dfrac{4 \times 48}{6.0 \times 10^{23} \times a^3}$

$a^3 = \dfrac{192}{30 \times 10^{23}} = 6.4 \times 10^{-23}\text{ cm}^3 = 64 \times 10^{-24}\text{ cm}^3$

Taking the cube root gives the edge length $a$:
$a = 4 \times 10^{-8}\text{ cm} = 4 \times 10^{-10}\text{ m}$

In a rock-salt lattice, the cations ($\text{M}^+$) form a face-centered cubic (FCC) sub-lattice. The shortest distance between two identical ions (next-nearest neighbors) is half the face diagonal of the unit cell:
Distance = $\dfrac{a\sqrt{2}}{2} = \dfrac{a}{\sqrt{2}}$

Substitute the value of $a$:
Distance = $\dfrac{4 \times 10^{-10}}{\sqrt{2}}\text{ m} = 2\sqrt{2} \times 10^{-10}\text{ m}$

Comparing this with the given expression $x\sqrt{2} \times 10^{-10}\text{ m}$, we get:
$x = 2$

Answer: $2$