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A metal of molar mass M g mol ⁻¹ crystallizes in a body-centered cubic (bcc) lattice. If the atomic radius of the metal is r pm , which of the following is the correct algebraic expression for the density of the metal in g cm ⁻³ ? (Here N_A is Avogadro's number)

Options

  1. A3 3 M 10²⁴ 32 N_A r^3
  2. B2 M 10³⁰ 8 N_A r^3
  3. C3 3 M 10³⁰ 32 N_A r^3
  4. D2 M 10³⁰ 16 N_A r^3

Correct answer

C. 3 3 M 10³⁰ 32 N_A r^3

Step-by-step solution

For a body-centered cubic (bcc) lattice, the number of atoms per unit cell is Z = 2 . The relationship between the edge length a and the atomic radius r for a bcc lattice is: 3 a = 4r a = 4r 3 Since the radius is given in picometers (pm), we must convert it to centimeters (cm) to find the density in g cm ⁻³ : a = 4r 3 10⁻¹⁰ cm The volume of the unit cell is: a^3 = ( 4r 3 10⁻¹⁰ )^3 = 64r^3 3 3 10⁻³⁰ cm ^3 The formula for density is: = Z M N_A a^3 Substitute the values of Z and a^3 into the density formula: = 2 M N_A

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