Naturally occurring boron is 80.20% boron-11 (of atomic mass 11.01 amu) and 19.80% of some other isotope form of boron. What must be the atomic mass of this second isotope in order to account for the 10.81 amu average atomic mass of boron?

A. 10.99 amu
B. 11.00 amu
C. None of these
D. 11.01 amu
E. 11.99 amu
F. 12.00 amu
G. 10.00 amu
H. 9.00 amu

0.8020(11.01) + 0.1980(x) = 10.81

Solve for x.

To solve this problem, we can use the concept of weighted average atomic mass. Here is how you can calculate the atomic mass of the second isotope of boron:

1. Let x be the atomic mass of the second isotope of boron.
2. Multiply the atomic mass of boron-11 by its percentage abundance: (11.01 amu) * (0.8020) = 8.81802 amu.
3. Multiply the atomic mass of the second isotope by its percentage abundance: (x) * (0.1980) = 0.198x amu.
4. Set up an equation by adding the two results from steps 2 and 3: 8.81802 amu + 0.198x amu = 10.81 amu.
5. Solve the equation for x by subtracting 8.81802 amu from both sides: 0.198x amu = 1.99198 amu.
6. Divide both sides of the equation by 0.198: x = 1.99198 amu / 0.198 = 10.06 amu.

Therefore, the atomic mass of the second isotope of boron must be 10.06 amu.

Since none of the answer options exactly match 10.06 amu, we can round it to the nearest available option. The nearest option is 10.00 amu, which is answer G.

So, the correct answer is option G.

X= 5.9089

Ans is 10.97amu