The natural boron of atomic weight 10.81 is found to have two isotopes 10 B and 11 B. The ratio of abundance of isotopes of natural boron should be:

Question

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

Tardigrade Admin · Accepted Answer

Let abundance of B 10 be x % ∴ Abundance of B 11(100- x ) % 10.81=((10 × x)+11(100-x)/100) or 1081=10 x+1100-11 x ⇒ x=19 ∴ Ratio of abundance =(19/100-19)=(19/81)

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

11: 10

81: 19

10: 11

19: 81

Let abundance of $B^{10}$ be $x %$
$∴$ Abundance of $B^{11} (100 - x) %$
$10.81 = \frac{( 10 \times x ) + 11 ( 100 - x )}{100}$
or $1081 = 10 x + 1100 - 11 x$
$\Rightarrow x = 19$
$∴$ Ratio of abundance
$= \frac{19}{100 - 19} = \frac{19}{81}$

Q. The natural boron of atomic weight 10.81 is found to have two isotopes 10B and 11B. The ratio of abundance of isotopes of natural boron should be: