Chemistry, asked by appasablonari, 3 months ago

silver from a a CCP letters the the edge length of its unit cell is 408.6 picometre calculate the density of a silver NA= 6.022× 10restiu 23 mol_1 atomic mass 108 g/ mol​

Answers

Answered by bibifathima475
1

Answer:

The edge length of it's unit cel is 408.6 picometre

or

408.6 × 10^-10cm

The volume of the unit cel =a^3

=(408.6 × 10^-10cm)^3

= 68.27 × 10^-24 cm^3

ccp unit cells has 4 atoms per unit cell

Mass of one Ag atom

= molar mass of Ag /a vogadros number

= 108g/mol /6.022× 10^23 /mol

= 17.96×10^23g

Mass of unit cell =4×17.93 ×10^23g

=71.72×10^23g

Density of silver = mass of unit cell/volume of unit cell

Density of silver = 71.72×10^23g/68.27×10^-24cm^3

= 10.51g/cm^3.

I hope it's help a lot.

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