CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Naturally occurring Boron consists of two isotopes whose atomic weights are 10.01 and 11.01. The atomic weight of natural Boron is 10.81. Calculate the percentage of each isotope in natural boron.


Open in App
Solution

Step 1: Assuming the percentage of the isotopes:

Let x% is the percentage of the isotope having the atomic weight = 10.01

So, (100-x)% would be the percentage of the isotope having the atomic weight =11.01

Step 2: Finding the percentage of each isotope:

Therefore,

Averageatomicweight=[(atomicweightofthefirstisotope)×(x)+(atomicweightofsecondisotope)×(100-x)]100

10.81=10.01(x)+11.01×(100-x)100

So, value of x = 20

Hence,

Percentage of isotope having the atomic weight of 10.01 = 20

and, Percentage of isotope having the atomic weight of 11.01 = (100 - 20) = 80


flag
Suggest Corrections
thumbs-up
21
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Empirical and Molecular Formula
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon