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The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 700°C is 2 × 10-6. Calculate the number of vacancies (per meter cubed) at 700°C. Assume a density of 10.35 g/cm3 for Ag, and note that AAg = 107.87 g/mol.

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olasunkanmilesanmi

Answer:

[tex]1.16*10^{23} vacancies/m^3[/tex]

Explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

[tex]\frac{N_v}{N}=2*10^{-6}\\N_v=2*10^{-6}N\\[/tex]

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

[tex]N=\frac{p_{Ag} * N_A}{A_{AG}} \\N=\frac{6.022*10^{23} * 10.35*10^6g/m}{107.87g/mol} \\N=5.78*10^{28}atom/m^3[/tex]

if we substitute the value of the number of lattice into the first equation, we arrive at

[tex]N_v=2*10^{-6} *5.78*10^28\\N_v=1.16*10^{23} vacancies/m^3[/tex]

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