Atomic weight of boron is 10.81 and it has two isotopes B105 and B115. Then ratio of B105:B115  in nature would be 
(a) 19 : 81                  (b) 10 : 11
(c) 15 : 16                  (d) 81 : 19

(a) Let the percentage of B10 atoms be x, then Average atomic weight

=10x+11(100-x)100=10.81x=19      NB10NB11=1981

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