Question

Boron found in nature having atomic weight of 10.811 and is made up of the isotopes $B^{10}$ (mass 10.013 amu) and isotope $B^{11}$ (mass 11.0093). What percentage of naturally occurring boron is made up of $B^{10}$ and $B^{11}$ respectively?

• A. 30:70
• B. 25:75
• C. 20:80
• D. 15:85
• E. 10:90

Answer 1

Answer: (C)

20:80

Let, P = percentage.
Atomic weight of Boron$=10.811=M_1\times \frac{P}{100}+M_2\times \left(\frac{100-P}{100}\right)=10\times \frac{P}{100}+11\times \left(\frac{100-P}{100}\right)$
$10.811=.\frac{10P}{100}+\frac{1100-11P}{100}$
$10.811=\frac{1100-P}{100}$
$1100-P=1081.1$
$P=1100-1081.1=18.9$%.
Hence, percentage of $B_{10}=18.9$%,
Percentage of $B_{11}=100-18.9=81.1$%.
As this ratio ( 18.9 : 81.1 ) is nearby 20 : 80, hence, answer is 20 : 80.