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Boron naturally occurs in two isotopic forms, 10B and 11B. If the atomic mass 10.81 amu what is the approximate ratio of 10B to 11B?Group of answer choices19:8135:6581:1925:7565:35

Question

Boron naturally occurs in two isotopic forms, 10B and 11B. If the atomic mass 10.81 amu what is the approximate ratio of 10B to 11B?Group of answer choices19:8135:6581:1925:7565:35

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Solution

The atomic mass of an element is the weighted average of the masses of its isotopes. The atomic mass of Boron is given as 10.81 amu.

Let's denote the natural abundance of Boron-10 as x and that of Boron-11 as (1-x).

The atomic mass of Boron-10 is 10 amu and that of Boron-11 is 11 amu.

We can set up the following equation based on the definition of atomic mass:

10x + 11(1-x) = 10.81

Solving this equation for x gives us the natural abundance of Boron-10.

10x + 11 - 11x = 10.81 -x = 10.81 - 11 -x = -0.19 x = 0.19

So, the natural abundance of Boron-10 is 0.19 or 19% and that of Boron-11 is 81%.

Therefore, the approximate ratio of Boron-10 to Boron-11 is 19:81.

This problem has been solved

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