Question

silver crystallize 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

Correct question :

Silver crystallize in fcc lattice. If edge length of the cell is $\rm{4.077\times{}10^{-8}\:cm}$ and density is $\rm{10.5\:g\:cm^{-3}}$.Calculate the atomic mass of silver.

Answer :

• Atomic mass of silver = $\rm{107.12\:g\:mol^{-1}}$

Step-by-step explanation :

Given that,

• Edge length, a = $\rm{4.077\times{}10^{-8}\:cm}$
• Density, ρ = $\rm{10.5\:g\:cm^{-3}}$

Also,

• Avogadro's number, $\rm{N_0}$ = $\rm{6.022\times{}10^{23}\:mol^{-1}}$

• For fcc lattice, Z = 4.

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We know, $\rm{Density,\:\rho=\dfrac{Z\times{}M}{a^3\times{}N_0}}$

$\bold{OR}$

$\rm{M=\dfrac{\rho\times{}a^3\times{}N_0}{Z}}$

Putting the given values in the expression, we get,

$\implies\rm{M=\dfrac{10.5\:g\:cm^{-3}\times{}(4.077\times{}10^{-8}\:cm)^3\times{}6.022\times{}10^{23}\:mol^{-1}}{4}}$

Solving, we get,

$\implies\rm{M=}\bold{107.12\:g\:mol^{-1}}$

Answer 2

From the equation, density = ZxM/a3 x N0, we get

M = ρ x a3 x N0/Z …………………………………1

Placing the values in eqatation 1,

WE GET M= 10.5 x (4.077 x10-8)3 x 6.022 x 1023/4

= 107.1 gmol-1

= Therefore, atomic mass of silver = 107.1 gmol-1