The equilibrium fraction of lattice sites that are vacant in silver (ag) at 500â°c is 0.5 ã 10-6. calculate the number of vacancies (per meter cubed) at 500â°c. assume a density of 10.35 g/cm3 for ag, and note that aag = 107.87 g/mol.

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 ×

um i think it might be a car?