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An element has atomic mass 93 g/mol and density 11-5 g/cmº. If edge length of itsunit cell is 300 p.m. identify the types of unit cell. |
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Answer» Number of atoms per unit cell, z = d × a³ × NA / M --> (1) Here, d = 11.5 g cm⁻³ , M = 93 g mol⁻¹ , NA = 6.022 × 10²³ mol⁻¹ a = 300 pm = 300 × 10⁻¹⁰ cm = 3 × 10⁻⁸ cm Now, Substituting these values in the expression (1), we get z = 11.5 g cm⁻³ × (3 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹ / 93 g mol⁻¹ FINAL RESULT = 2.01 As there are 2 atoms of the element present unit cell, therefore, the cubic unit cell must be body centered |
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