Question

average atomic weight of boron is 10.10 and boron exists in two isotopic forms B10 And B 11 The percentage abundance of B 10 is
1.10
2.90
3.50
4.20
Please give the full solution☺

Answer 1

Let % abundant of B10 be x

Therefore, % abundant of B11 = 100 - x

Avr. at. wt. = (x)10 + (100 - x)11 / 100

=> 1010 = 10x + 1100 -11x

=> x = 90%

Answer 2

Answer:

90 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, B 10:

% = x %

Mass = 10 u

For second isotope, B 11:

% = 100  - x  

Let, Mass = 11 u

Given, Average Mass = 10.10 u

Thus,  

$10.10=\frac {x}{100}\times {10}+\frac {100-x}{100}\times {11}$

Solving for x, we get that:

x = 90 %

Thus percentage abundance of B 10 is = 90 %