Silver chloride (AgCl) adopts the rock salt structure. The density of \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\) is $5.56 \mathrm{~g} / \mathrm{cm}^{3}$. Calculate the length of an edge of the AgCl unit cell.

To find the molecular weight of AgCl, we first need to know the atomic weights of silver (Ag) and chlorine (Cl). The atomic weight of Ag is 107.87 g/mol, and the atomic weight of Cl is 35.45 g/mol. Therefore, the molecular weight of AgCl is the sum of the two atomic weights: Molecular weight of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

In the rock salt structure, each metal atom (Ag) is surrounded by six halide atoms (Cl), and each halide atom is surrounded by six metal atoms. This means there is one Ag atom and one Cl atom in each unit cell. Therefore, there are two AgCl units per unit cell.

The formula for the density (ρ) of a unit cell is: ρ = (mass of one unit cell) / (volume of one unit cell) We can also write this formula in terms of the number of AgCl units, molecular weight, and edge length: ρ = (number of AgCl units × molecular weight of AgCl) / (volume of one unit cell) Since the volume of a cubic unit cell is equal to the cube of its edge length (a^3), we can write: ρ = (number of AgCl units × molecular weight of AgCl) / (edge length^3) Now we can plug in the given density (5.56 g/cm³), number of AgCl units (2 units), and molecular weight of AgCl (143.32 g/mol), and solve for the edge length (a): 5.56 g/cm³ = (2 × 143.32 g/mol) / (edge length^3) To find the edge length, first, divide both sides by the molecular weight of AgCl(286.64 g/mol): \( \frac{5.56 g/cm^3}{286.64 g/mol} = \frac{1}{a^3} \) Now, calculate the cube root of both sides to find the edge length: \( a = \sqrt[3]{\frac{286.64 g/mol}{5.56 g/cm^3}} \) Using a calculator, we obtain: a ≈ 5.63 Å

Since we've found the edge length (a) in Ångström (Å), we can express it in the appropriate units. Recall that 1 Å = 10⁻¹⁰ meters. Edge length of the AgCl unit cell ≈ 5.63 Å ≈ 5.63 x 10⁻¹⁰ meters. Given that the edge length of the AgCl unit cell is approximately 5.63 Å, the response to the problem is complete.