Question

Silver crystallises in fcc structure with edge length of unit cell, 4.07 x 10⁻⁸ cm and if density of metallic silver is 10.5 g cm⁻³. Calculate the molecular mass of silver. (107.8)

Answer 1

Answer:

A= 4.07 × 10^-8

d= 10.5 gm / cm3

Z= 4

m= ?

Explanation:

d= (Z×m)÷ (n×a3)

m = {10.5× 6.023× 10^ 23× (4.07 × 10^-8)^3} ÷ 4

= 106.592 gm

Answer 2

Solution :

 Given ,

 Edge Length (a)=4.07× 10⁻⁸ cm

 density (ρ) = 10.5 g cm⁻³

 For, fcc number of atoms per unit cell (z)= 4

 We know that ,

 Atomic Mass (M) = ρ×a³×Na/z



M =10.5 g cm⁻³×(4.07×10⁻⁸cm)³×6.022×10²³ / 4

M =10.5 g cm⁻³×67.419×10⁻²⁴cm³×6.022×10²³/4

M =10.5 g ×67.419×10⁻¹×6.022mol⁻¹ /4

M =10.5 g ×40.599mol⁻¹ /4

M =426.2895 g mol⁻¹ /4

M =106.572g mol⁻¹ =107g mol⁻¹ approx.

Hence, atomic mass of silver = 107g mol⁻¹

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