Silver crystallises in fcc lattice- If edge length of the cell is 4-07 - 10-8 cm and density is 10-5 g cm -3- calculate the atomic mass of silver-

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 kno ...

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Question

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g cm⁻³, calculate the atomic mass of silver.

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Solution

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, N_A = 6.022 × 10²³ mol⁻¹

Using the relation:

= 107.13 gmol⁻¹

Therefore, atomic mass of silver = 107.13 u

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Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver.

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: = 107.13 gmol−1 Therefore, atomic mass of silver = 107.13 u

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g cm⁻³, calculate the atomic mass of silver.

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, N_A = 6.022 × 10²³ mol⁻¹

Using the relation:

= 107.13 gmol⁻¹

Therefore, atomic mass of silver = 107.13 u