There are two isotopes of boron: boron-10 and boron-11. explain, in terms of weighted average atomic mass, the difference in percent abundance of each of the isotopes of boron.

Percent abundance of  B¹⁰ is 19.9% and  B¹¹ 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 B¹⁰ (10.012938) and B¹¹ (11.009305)

We will determine the percent abundance of each isotopes.

First of all we will set the fraction for both isotopes B¹⁰ and  B¹¹.

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 +  10.012938  ( 10.012938 -x) = 10.811

11.009305x +  10.012938 -  10.012938 x = 10.811

11.009305x   -  10.012938 x = 10.811 -  10.012938

0.996367x = 0.798062

x= 0.798062 / 0.996367

x= 0.800972

Percent abundance of B¹¹.

0.800972 × 100 = 80.1 %

80.1 % is abundance of B¹¹ because we solve the fraction x.

now we will calculate the abundance of B¹⁰.

(1-x)

1-0.800972 =0.199

Percent abundance of  B¹⁰.

0.199 × 100=  19.9%

19.9% for B¹⁰.

Now we can calculate the average atomic mass of boron.

Formula:

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

Now we will put the values in formula.

Average atomic mass = [19.9 × 10.012938] + [80.1 × 11.009305] / 100

Average atomic mass = 199.2575 + 881.8453 / 100

Average atomic mass = 10.811 amu

the answer is 0.8m.

by: nicholas

i wish i could but this one is weird give me abc and i can solve it.