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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.

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.

Grade:upto college level

3 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Sol. 1. Average atomic weight = (∑▒〖Percentage of an isotope ×atomic weight〗)/100 ⇒ 10.81 = (10.01x+11.01 (100-x))/100 ⇒ x = 20% Therefore, natural boron contain 20% (10.01) isotope and 80% other isotope.
jatin
11 Points
7 years ago
Let the % of boron with mass no 10 be A. And Mass no 11 be = 100 - A10.A+(100 - A) 11/100= 10.810A + 1100 -11A = 1080-1A = -20A=-20/-1A =20%And for boron with mass no.11=20-100 =80%
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
Let the % of boron with mass no 10.01 be x.
Percentage of Boron of mass no 11.01 = 100 – x
[10.01x + (100 – x) 11.01]/100 = 10.81
Hence, x = 20%
Percentage of Boron with mass no. 10.01 = 20%
And for boron with mass no.11.01 = 100 – 20 = 80%
 
Thanks and regards,
Kushagra

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