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.

Answer 1

Given,
edge length of the cell (a) = 4.07 × 10^-8 cm
density (d) = 10.5 g/cm³
Avogadro's number, N = 6.023 × 10²³ /mol
for fcc lattice, number of atoms per unit cell , z = 4

use formula $\bf{M=\frac{da^3N}{z}}$
= {10.5 × (4.07 × 10^-8)³ × 6.023 × 10²³}/4
= {10.5 × 67.419 × 0.6023 )/4
= 106.5 g/mol

hence, atomic mass of silver = 106.59 g/mol

Answer 2

Hey there!

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Answer :

$4Ca^{2+}$ and $8F^{-}$ ions means $4CaF_{2}$ units.

∴ Z = 4, d = $\frac{Z × M}{N_{A} × a^{3}}$

⇒ $N_{A} = \frac{Z × M}{d × a^{3}}$

⇒ $\frac{4 × 78.08 g mol^{-1}}{3.18 g cm^{-3} × (5.46)^3 × 10^{-24} cm^{3}}$

⇒ $\frac{312.32 × 10^{24}}{3.18 × (5.46)^3} = \frac{3123.2}{517.61}$ × $10^{23}$

⇒ $N_{A}$ = 6.0338 × $10^{23}$ $mol^{-1}$