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Naturally occurring Boron consists of two isotopes whose atomic masess are 10.01 and 11.01. The atomic mass of natural Boron is 10.81. Calculate the percentage of each isotope in natural Boron. |
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Answer» SOLUTION :Suppose the PERCENTAGE of isotope with atomic mass `10.01=x` Then percentage of isotope with atomic mass `11.01=100-x` `therefore" Average atomic mass"=(10.01x+(100-x)xx11.01)/(100)=(10.01x+1101-11.01x)/(100)=(1101-x)/(100)` `therefore""(1101-x)/(100)=10.81"(GIVEN)or"1101-x=1081"or"x=1101-1081=20` `therefore""%" AGE of isotope with atomic mass 10.01"=20%` `""%" age of isotope with atomic mass 11.01"=80%` |
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