silver forms a ccp lattice the edge lengthof its unit cell is 408.6pm calculate the densityof silver[NA = 6.022×10®23 atomic mass of ag= 108g mol®-1
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Answer:
10.5 g/cm^3
Explanation:
Let no. of atoms in a unit cell be " Z "
ccp lattice means that a unit cell contains 4 atoms
Therefore Z = 4
Given atomic mass of silver (M) = 108 g
Avagadro number (NA) = 6.022 × 10^23
Given edge length of unit cell (a) = 408.6 pm = 408.6 × 10^-10 cm
Therefore volume of unit cell = a^3 = (408.6 × 10^-10)^3 = 68.2 × 10^-24 cm^3
Formula for finding density = (Z × M) ÷ ( NA × volume )
= (4 × 108) ÷ (6.022 × 10^23 × 68.2 × 10^-24) g/cm^3
= 10.5 g/cm^3
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