Respuesta :
Answer:
The correct answer is option b.
Explanation:
Let the fractional abundance of B-10 be x
And fractional abundance of B-11 be (1- x)
Average atomic mass of boron = 10.81 amu
Atomic mas of B-10 = 10.0129 amu
Atomic mas of B-11 = 11.0093 amu
To determine an average atomic mass of an element we use::
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
[tex]10.81=10.0129 amu\times x+11.0093 amu\times (1- x)[/tex]
x = 0.200 = 0.200 × 100= 20%
(1-x) = 0.800 =0.800 × 100= 80%
The percent abundance of isotopes B-10 and B-11 is 20% and 80% respectively.
Considering the definition of atomic mass, isotopes and atomic mass of an element, the percent abundance of each isotope is 20% 10B and 80% 11B.
Definition of atomic mass
First of all, the atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.
Definition of isotope
The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.
Definition of atomic mass
On the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.
Percent abundance of each isotope
In this case, you know:
- Mass of 10B isotope: 10.0129 amu
- Percent abundance of 10B isotope: y
- Mass of 11B isotope: 11.0093 amu
- Percent abundance of 10B isotope: 1 - y
- Average atomic mass of boron= 10.81 amu
Then, the average mass of boron can be calculated as:
10.0129 amu×y + 11.0093 amu×(1-y)= 10.81 amu
Solving for "y":
10.0129 amu×y + 11.0093 amu×1- 11.0093 amu×y= 10.81 amu
10.0129 amu×y + 11.0093 amu- 11.0093 amu×y= 10.81 amu
10.0129 amu×y - 11.0093 amu×y= 10.81 amu - 11.0093 amu
(-0.9964 amu)×y= -0.1993 amu
y= (-0.1993 amu)÷ (-0.9964 amu)
y= 0.2= 20%
So:
- Percent abundance of 10B isotope: y= 0.2= 20%
- Percent abundance of 10B isotope: 1 - y= 1 - 0.2= 0.8= 80%
In summary, the correct answer is option (b): the percent abundance of each isotope is 20% 10B and 80% 11B.
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