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The Solid State

Question
CBSEENCH12005491

Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 x 10–8 cm and its density is 10.5 g cm-3. Calculate on this basis the atomic mass of silver. (N= 6.02 x 1023 mol–1).

Solution

We have,
Edge of length of cell a = 4.07x10–8cm
Density p = 10.5 g /cm3
Number of atoms in unit cell of fcc lattice = 4
Avogadro number NA = 6.022x1023
By using formula,
Density space equals straight p space equals fraction numerator ZM over denominator straight a cubed straight N subscript straight A end fraction
straight M space equals space fraction numerator Pa cubed straight N subscript straight A over denominator straight Z end fraction

Putting space the space value space in space the space above space equation space we space get comma

fraction numerator 10.5 space straight x left parenthesis 4.07 right parenthesis cubed space straight x 6.022 space space straight x 10 to the power of 23 over denominator 4 end fraction
solving space the space equation space we space get comma

straight M equals space 107.09 straight g space Mol to the power of negative 1 end exponent