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Amit Saxena Grade: upto college level
        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.
3 years ago

Answers : (2)

Deepak Patra
askIITians Faculty
474 Points
										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.
3 years ago
jatin
11 Points
										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%
										
9 months ago
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