JEE MainChemistrySolid State
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
- A3 3 M 10²⁴ 32 N_A r^3
- B2 M 10³⁰ 8 N_A r^3
- C3 3 M 10³⁰ 32 N_A r^3
- 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