Respuesta :
Density of ZnS given that the zn-s distance and bond angle are 0. 234 nm and 109. 5o, respectively if the atomic weights of zinc and sulfur are 65. 41 g/mol and 32. 06 g/mol. is 4.1109 g/cm³
For density In the crystal structure, we determine the angle's value for one set of ZnS bonds. θ θ + ∅ + 90° = 180° θ = 90° - ∅ θ = 90° - (109.5° / 2) After determining that = 35.25°, we derive the value of x from the geometry: distance angle d = 0.234 x = dsin = 0.234 sin35.25°) = 0.135 nm = 0.135 107 cm; following this, we get the length of the unit cell: a = 4x a = 4(0.135) a = 0.54 nm = 0.54 107 cm; and The formula for n' is (number of corner atoms in unit ell contribution of each corner atom in unit cell) + (number of face center atoms in unit cell contribution of each face center atom in unit cell). n' = 8 × 1/8) + (6 × 1/2) = 1 + 3 = 4
Density = P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)] = 389.88 / 94.84 = 4.1109 g/cm³
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