The natural boron of atomic weight 10.81 is found to have two isotopes B10 and B11 . The ratio of abundance of isotopes in natural boron should be

Question

The natural boron of atomic weight 10.81 is found to have two isotopes B10 and B11 . The ratio of abundance of isotopes in natural boron should be (A)  11:10  (B)  81:19  (C)  10:11  (D)  15:16

Tardigrade Admin · Accepted Answer

Let abundance of B10 be m% . So, abundance of B11=(100-m)% ∴ 10.81=((10× m)+11(100-m)/100) Or 1081=10m+1100-11m Or m=19 ∴ Ratio of abundances =(19/100-19)=(19/81)

Q. The natural boron of atomic weight $10.81$ is found to have two isotopes $B^{10}$ and $B^{11}$ . The ratio of abundance of isotopes in natural boron should be

$11 : 10$

$81 : 19$

$10 : 11$

$15 : 16$

$19 : 81$

Let abundance of $B^{10}$ be m% .
So, abundance of $B^{11} = (100 - m)$
$∴$ $10.81 = \frac{( 10 \times m ) + 11 ( 100 - m )}{100}$
Or $1081 = 10 m + 1100 - 11 m$
Or $m = 19$
$∴$ Ratio of abundances
$= \frac{19}{100 - 19} = \frac{19}{81}$

Q. The natural boron of atomic weight 10.81 is found to have two isotopes B10 and B11 . The ratio of abundance of isotopes in natural boron should be

11:10

81:19

10:11

15:16

19:81

Let abundance of B10 be m% . So, abundance of B11=(100−m) ∴ 10.81=100(10×m)+11(100−m)​ Or 1081=10m+1100−11m Or m=19 ∴ Ratio of abundances =100−1919​=8119​