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

Boron found in nature has an atomic weight of 10.811 and is made up of the isotopes 10B (mass 10.013 amu) and 11B (mass 11.0093 amu). What percentage of naturally occurring boron is made up of 10B and 11B respectively?

A
30:70
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
25:75
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
20:80
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
15:85
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 20:80
Let us consider in boron, B10 and B11 isotopes are present in the ratio of 20:80.
B10 has a mass of 10.013amu. Thus 20% of it will have a weight of 20100×10.013, that is 2.0026.
B11 has a mass of 11.0093amu. Thus 80% of it will have a mass of 80100×11.0093, that is 8.8074
Thus, total weight of boron which consists of B10 and B11 isotopes is 2.0026+8.8074, which is equal to 10.81, which matches with our given question.
Thus, by back calculation, we find that 20:80 gives the correct ratio of isotopes.
Option C is the correct ratio.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Refining
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon