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Since KCl has the same structure as NaCl, it is a face-centered cubic (fcc) lattice. In an fcc lattice, the atoms are located at the corners and the centers of the faces of the cube. The smallest distance between ions in KCl is given as 314 pm.

In an fcc lattice, the shortest distance between ions lies along the face diagonal. Let 'a' be the edge length of the unit cell, and 'd' be the shortest distance between the ions. Using Pythagorean theorem for right triangles, we can write: \(a^2 + a^2 = d^2\) Since there are two K+ ions and two Cl- ions lying along the face diagonal: \(2a^2 = d^2\) Now we can use the given shortest distance (d = 314 pm) to calculate the edge length (a): \(2a^2 = (314 \, \text{pm})^2\) Solve for 'a': \(a^2 = \frac{(314 \, \text{pm})^2}{2}\) \(a = \sqrt{\frac{(314 \, \text{pm})^2}{2}}\) \(a \approx 222 \, \text{pm}\)

To calculate the density of KCl, we need to find the mass of KCl in the unit cell and divide it by the volume of the unit cell. In an fcc lattice, there are 4 formula units of KCl per unit cell (8 corner ions and 6 face-centered ions with each of them contributing 1/2). The molar mass of KCl is approximately 39.1 g/mol (for K+) and 35.5 g/mol (for Cl-). Therefore, the mass of 4 formula units of KCl is: \(m_{\text{KCl}} = 4 \times (39.1 \, \text{g/mol} + 35.5 \, \text{g/mol}) = 298.4 \, \text{g/mol}\) Now we can find the volume of the unit cell by using the cube formula and the edge length: \(V = a^3 \approx (222 \, \text{pm})^3 = 1.093 \times 10^{-29} \, \text{m}^3\) Finally, we can find the density (蟻) of KCl by dividing the mass by the volume: \(\rho = \frac{m_{\text{KCl}}}{V} = \frac{298.4 \, \text{g/mol}}{1.093 \times 10^{-29} \, \text{m}^3}\) \(\rho \approx 2.73 \times 10^6 \, \text{g/m}^3\) So, the density of KCl is approximately 2.73 x 10^6 g/m鲁.