Answer:
Percent abundance of Ā BĀ¹ā° is 19.9% and Ā BĀ¹Ā¹ is 80.1%.
The weighted average atomic mass of boron is 10.811 amu.
Explanation:
We know that average atomic mass of Boron is 10.811 amu and there are two isotopes of boron BĀ¹ā° (10.012938) and BĀ¹Ā¹ (11.009305)
We will determine the percent abundance of each isotopes.
First of all we will set the fraction for both isotopes BĀ¹ā° and Ā BĀ¹Ā¹.
X for the isotopes having mass 11.009305 amu.
1-x for isotopes having mass 10.012938 amu.
The average atomic mass of Boron is 10.811 amu
we will use the following equation,
11.009305x + Ā 10.012938 Ā ( 10.012938 -x) = 10.811
11.009305x + Ā 10.012938 - Ā 10.012938 x = 10.811
11.009305x Ā - Ā 10.012938 x = 10.811 - Ā 10.012938
0.996367x = 0.798062
x= 0.798062 / 0.996367
x= 0.800972
Percent abundance of BĀ¹Ā¹.
0.800972 Ć 100 = 80.1 %
80.1 % is abundance of BĀ¹Ā¹ because we solve the fraction x.
now we will calculate the abundance of BĀ¹ā°.
(1-x)
1-0.800972 =0.199
Percent abundance of Ā BĀ¹ā°.
0.199 Ć 100= Ā 19.9%
19.9% for BĀ¹ā°.
Now we can calculate the average atomic mass of boron.
Formula:
Average atomic mass = [mass of isotopeĆ its abundance] + [mass of isotopeĆ its abundance] +...[ ] / 100
Now we will put the values in formula.
Average atomic mass = [19.9 Ć 10.012938] + [80.1 Ć 11.009305] / 100
Average atomic mass = 199.2575 + 881.8453 / 100
Average atomic mass = 10.811 amu
