Ordinary boron is a mixture of 10B5(mass no=10,atomic no=5)&11B5(mass no=11,atomic no=5)isotopes & has a composite atomic mass of 10.82u.What percentage of each isotope is present in ordinary boron?
1 Answer
May 18, 2018
Explanation:
Solves as:

#p = 9 /50 equiv 18"%" # 
# implies 1  p = 41/50 equiv 82 "%"#