Question

24 An element with density 11.2 g em forms a f.c.e. lattice with edge length of
4 x 10 cm. Calculate the atomic mass of the element (Given: N. = 6.022 x
10” mol)
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Answer 1

Answer:

M = 108

Explanation:

Density d=11.2g/cm

3

Edge length a=4×10⁸cm

Avogadro's number=6.023×10²³mol

Atomic mass M=??

For f.c.c lattice, number of atoms per unit cell, Z=4

d=N×a³/m ×Z

M= d ×N ×a³/z

M= 4×11.2×6.023×10²³×(4×10⁸)³

M=108g/mol

Hence, the atomic mass of the element is 108g/mol. It is silver metal.