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  • Silver crystallises in fcc lattice If edge length of the cell is 4.07x10 to the power of -8 cm and density is 10.5 g cm to the power of -3,calculate the atomic mass of silver.

1.11   Silver crystallises in fcc lattice. If edge length of the cell is 4.07\times 10^{-8}cm and density is 10.5\; g\; cm^{-3} , calculate the atomic mass of silver.

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The relation between density, molar mass and edge length is given by:-

                                                 Density = \frac{z.M}{a^3.N_A}

It is given that silver crystallises in fcc lattice, so the value of z = 4. (since it is known that 4 atoms are completely efficient in fcc lattice)                 

                                                          M = \frac{Density.a^3.N_A}{z} 

                                                              = \frac{10.5\ g\ cm^{-3} \times (4.07\times10^{-8}cm)^3\times N_A}{4}

 Convert these into SI units,                = 107.09\ u

                                                               

Posted by

Devendra Khairwa

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