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Chapter 12: Problem 60

Clausthalite is a mineral composed of lead selenide (PbSe). The mineral adopts the rock salt structure. The density of PbSe at \(25^{\circ} \mathrm{C}\) is 8.27 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate the length of an edge of the PbSe unit cell.

Short Answer

Expert verified
The length of an edge of the PbSe unit cell is approximately 6.05 x 10^-8 cm.

Step by step solution

01

Determine the number of atoms per unit cell

In the rock salt structure, each unit cell is occupied by four formula units (2 for lead and 2 for selenium). So there are 4 atoms of PbSe per unit cell.
02

Calculate the molar mass of PbSe

To calculate the molar mass of PbSe, we first find the molar masses of lead (Pb) and selenium (Se) and then add them together. Molar mass of Pb: 207.2 g/mol Molar mass of Se: 78.96 g/mol Molar mass of PbSe = Molar mass of Pb + Molar mass of Se = 207.2 + 78.96 = 286.16 g/mol
03

Find the mass of atoms per unit cell

Now we need to find the mass of atoms in one unit cell. Since there are 4 atoms of PbSe per unit cell, the mass of atoms in one unit cell can be calculated as: Mass of atoms per unit cell = (4 atoms/unit cell) x (Molar mass of PbSe / Avogadro's number) = 4 x (286.16 g/mol) / (6.022 x 10^23 atoms/mol) The mass of atoms per unit cell = 1.904 x 10^-22 g
04

Use the formula for the density of a crystal lattice

The formula for the density (ρ) of a crystal lattice is given by: ρ = (Mass of atoms per unit cell) / (Volume of unit cell) We're given the density, ρ = 8.27 g/cm³, and we know the mass of atoms per unit cell. The volume of a unit cell can be expressed as the cube of the length of its edge (V = a^3, where a is the length of an edge of the unit cell). So, we can rewrite the formula for density as follows: ρ = (Mass of atoms per unit cell) / (a^3) Rearranging to find a: a^3 = (Mass of atoms per unit cell) / ρ a = ((1.904 x 10^-22 g) / (8.27 g/cm³))^(1/3)
05

Calculate the length of an edge of the PbSe unit cell

Now we have all the required information to calculate the length of an edge of the PbSe unit cell. Plugging in the values into the formula, we get: a = ((1.904 x 10^-22 g) / (8.27 g/cm³))^(1/3) ≈ 6.05 x 10^-8 cm Thus, the length of an edge of the PbSe unit cell is approximately 6.05 x 10^-8 cm.

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