Answer:
Percent yield = 94.62%
Explanation:
Given data:
Actual yield of BeBr₂ = 1.23 mol
Theoretical yield of BeBr₂ = 220 g
Percent yield = ?
Solution:
Theoretical yield in moles:
Number of moles = mass/molar mass
Number of moles = 220 g/168.82 g/mol
Number of moles = 1.30 mol
Percent yield:
Percent tiled = actual yield / theoretical yield × 100
by putting values,
Percent yield = 1.23 mol / 1.30 mol × 100
Percent yield = 0.9462 × 100
Percent yield = 94.62%
Hey there!
The number of vacancies per unit volume => ( Nv = 2.3*10²⁵ m⁻³ )
Avogrado's number => ( NA = 6.022*10²³ atoms/mol )
Density of material ( p ) in g/m³ :
1 g/cm³ = 1000000 g/m³ so:
7.40 * ( 1000000 ) = 7.40*10⁶ g/m³
Atomic mass = 85.5 g/mol
* Calculate the number of atomic sites per unit volume :
N = NA * p / A
N = ( 6.022*10²³ ) * ( 7.40*10⁶ ) / 85.5
N = 4.45*10³⁰ / 85.5
N = 5.212*10²⁸ atoms/m³
Therefore:
Calculate the fraction of vacancies :
Fv = Nv / N
Fv = 2.3*10²⁵ / 5.212*10²⁸
FV = 4.441*10⁻⁴
Hope that helps!