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Sagot :
In order to determine the number of formula units of a substance in a cell, you need to know the density of the substance and the edge length of the unit cell. Suppose that you are given the following problem:The density of TlCl(s) is 7.00 g/cm³, and the length of an edge of a unit cell is 385 pm. How many formula units of TlCl are in a unit cell?
Here’s how you solve the problem.Calculate the volume of the unit cell.
If the length of an edge of the cell is a, then V = a³.
V = (385 × 10⁻¹² m)³ = 5.71 × 10⁻²⁹ m³Use the density to calculate the mass of the unit cell (The density of TlCl is the same, no matter what volume of TlCl we have).
Mass = 5.71 × 10⁻²⁹ m³ × (7.00 g/1 cm³) × (100 cm/1 m)³ = 3.99 × 10⁻²² gUse the molar mass to convert grams to moles.
3.99 × 10⁻²² g × (1 mol/239.8 g) = 1.67 × 10 ⁻²⁴ molUse Avogadro’s number to calculate the number of formula units.
1.67 × 10 ⁻²⁴ mol × (6.022 × 10²³ formula units/1 mol) = 1.00 formula units
Therefore, 1 unit cell of TlCl contains 1 formula unit.
Here’s how you solve the problem.Calculate the volume of the unit cell.
If the length of an edge of the cell is a, then V = a³.
V = (385 × 10⁻¹² m)³ = 5.71 × 10⁻²⁹ m³Use the density to calculate the mass of the unit cell (The density of TlCl is the same, no matter what volume of TlCl we have).
Mass = 5.71 × 10⁻²⁹ m³ × (7.00 g/1 cm³) × (100 cm/1 m)³ = 3.99 × 10⁻²² gUse the molar mass to convert grams to moles.
3.99 × 10⁻²² g × (1 mol/239.8 g) = 1.67 × 10 ⁻²⁴ molUse Avogadro’s number to calculate the number of formula units.
1.67 × 10 ⁻²⁴ mol × (6.022 × 10²³ formula units/1 mol) = 1.00 formula units
Therefore, 1 unit cell of TlCl contains 1 formula unit.
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