Question

An element crystallizes in foc lattice. If edge length of the unit cell is 408.6 pm and
density is 10,5 g cm". Calculate the atomic mass of the element (NA = 6.022 x 1023).​

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An element crystallises in fcc lattice. If edge length of the unit cell is

An element crystallises in fcc lattice. If edge length of the unit cell is 408.6 pm and density is 10.5 g cm¯³, calculate the atomic mass of the element [ N a = 6.022 * 10²³]

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A)

It is given that the edge length, a = 4.077 × 10−8 cm Density,

d = 10.5 g cm−3

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that

NA = 6.022 × 1023 mol−1 Using the relation:

d=zMa3NA

M=da3NAz

10.5gcm−1×(4.077×10−8cm)3×6.023×1023mol−14

=107.13gmol−1

Hence the atomic mass of silver is 107.13 u