An element with a molar mass of 81 g mol ⁻¹ crystallises in a body-centred cubic (bcc) lattice. The density of the crystal is 10 g cm ⁻³ . The atomic radius of the element in pm is (Nearest integer) [Given: N_A = 6 10²³ mol ⁻¹ , 3 = 1.73 ]

Step-by-step solution

The density of a unit cell is given by: d = Z M N_A a³ For a bcc lattice, the number of atoms per unit cell ( Z ) is 2 . Rearranging the formula to solve for the volume of the unit cell ( a³ ): a³ = Z M d N_A a³ = 2 81 10 6 10²³ = 162 60 10²³ a³ = 2.7 10⁻²³ cm ³ = 27 10⁻²⁴ cm ³ Taking the cube root, the edge length a is: a = 3 10⁻⁸ cm = 300 pm For a bcc lattice, the relationship between the atomic radius ( r ) and the edge length ( a ) is: r = 3 4 a r = 1.73 4 300 = 1.73 75 = 129.75 pm Rounding to the nearest integ