Respuesta :
Answer:
The answer to your question is :
a) Abundance of Li-6 = 7.49
b) Abundance of Li-7 = 100-7.49 = 92.5
Explanation:
Data Isotopes Mass Abundance
Li-6 6.015 amu y
Li-7 7.016 amu 100 -y
Lithium atomic mass = 6.941
6.941 = (6.015y) + (7.016)(100-y) /100
6.941 = (6.015y + 701.6 - 7.016y) / 100
6.941 x 100 = 6.015y - 7.016y + 701.6
694.1 -701.6 = - 1.001y
-7.5 = -1.001 y
y = 7.49
Abundance of Li-6 = 7.49
Abundance of Li-7 = 100-7.49 = 92.5
Answer:
a) The relative abundance of Li-6 =0.0749 = 7.49%
The relative abundance of Li-7 =1 - 0.0749 = 0.9251 =92.51%
b) The fraction of sample of lithium atoms with exact mass of 6.941 amu is 0.
Explanation:
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]....(1)
Let the fractional abundance of Li-6 isotope be 'x'. So, fractional abundance of Li-7 isotope will be '1 - x'
For Li-6 isotope:
Mass of Li-6 isotope = 6.015 amu
For Li-7 isotope:
Mass of Li-7 isotope = 7.016 amu
Average atomic mass of lithium = 6.941 amu
[tex]6.914 amu=x\times 6.015 amu+(1-x)\times 7.016 amu[/tex]
x= 0.0749
a) The relative abundance of Li-6 =0.0749 = 7.49%
The relative abundance of Li-7 =1 - 0.0749 = 0.9251 =92.51%
b) The fraction of sample of lithium atoms with exact mass of 6.941 amu is 0. This is because 7.49% of the sample has atomic mass of 6.015 amu and 92.51 % of sample has mass of 7.016 amu. Either the sample atom of lithium will have mass of 6.015 amu or 7.016 amu but it will not be having mass of exact 6.941
