Respuesta :
Answer:
Percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 80 %
Explanation:
The formula for the calculation of the average atomic mass is:
[tex]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})[/tex]
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 = 10.0129 u
For second isotope, B 11:
% = 100 - x
Mass = 11.0093 u
Given, Average Mass = 10.81 u
Thus,
[tex]10.81=\frac{x}{100}\times {10.0129}+\frac{100-x}{100}\times {11.0093}[/tex]
[tex]10.0129x+11.0093\left(100-x\right)=1081[/tex]
Solving for x, we get that:
x = 20 %
Thus percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 100 - 20 % = 80 %
Answer:
The relative abundance of 10B = 0.20 (20%)
The relative abundance of 11B = 0.80 (80%)
Explanation:
Step 1: Data given
Boron has 2 natural isotopes
⇒ 10B has a mass of 10.0129 u
⇒ 11B has a mass of 11.0093 u
Average atomic mas of Boron = 10.81
Step 2:
10B has an abundance of X %
11B has an abundance of Y %
X+ Y = 1
X = 1 - Y
10.81 = 10.0129* (1 - Y) + 11.0093*Y
10.81 = 10.0129 - 10.0129Y + 11.0093Y
0.7971 = 0.9964Y
Y = 0.80
X = 1.0 - 0.80 = 0.20
10.0129*0.20 + 11.0093*0.80 = 2.00258 +8.8044 = 10.81
The relative abundance of 10B = 0.20 (20%)
The relative abundance of 11B = 0.80 (80%)
