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Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u). |
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Answer» Solution : Since the lattice is ccp, the number of Ag atoms per unit cell (z) = 4 Molar mass of Ag `= 107.9 "g mol"^(-1)` Edge length of (a) = 408.6 PM Volume of cube = `(408.6)^3 xx 10^(-30) "CM"^3` `d = (Z xx M)/(N_A xx a^3)` `therefore d= (4 xx (107.9 "g mol"^(-1)))/((6.022 xx 10^(23) "atomsmol"^(-1)) xx (408.6)^3 xx 10^(-30) "cm"^3)` `therefore d = 10.5 "g cm"^(-3).` |
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