Answer : The number of vacancies (per meter cube) = 2.889 ×
Solution : Given,
Atomic weight of silver (Ag), = 107.87 g/mol
Density of silver (Ag), = 10.35 = 10.35 × = 10.35 ×
Avogadro's number, = 6.022 × atoms/mol
Fraction of lattice sites that are vacant in silver = 0.5 ×
Formula used :
Where,
= Total number of lattice sites in Ag
= Avogadro's number
= Density of silver
= Atomic weight of silver
This problem is solved by two steps:
step 1 : First we calculate the total number of lattice sites in silver.
Now put all the values in above formula, we get
= 5.778 × atoms/
Step 2 : Now multiply the value of by the fraction of lattice sites that are vacant in silver, 0.5 × , we get the number of vacancies (per meter cube)
The number of vacancies (per meter cube) = 5.778 × atoms/ × 0.5 × = Answer : The number of vacancies (per meter cube) = 2.889 ×