Question

The percentage of two stable isotopes ${}_{\text{5}}{{\text{B}}^{{\text{11}}}}$ and ${}_{\text{5}}{{\text{B}}^{{\text{10}}}}$ is, given that average atomic mass of boron is ${\text{10}}{\text{.81amu}}$ .
A.$80.1\% $ and $19.9\% $
B.$82.1\% $ and $17.9\% $
C.$81.0\% $ and $19.0\% $
D.$79\% $ and $21\% $

Answer 1

Hint: When two atoms of the same element possess different numbers of neutrons, they have different masses and they are called isotopes of that element.
Isotopes occur naturally and so the percent abundance of two isotopes in nature can be calculated if the atomic masses and the element’s average atomic masses are known.

Complete step by step answer:
The percent abundance of the isotopes of an element can be determined by using the equation:
${\text{a = }}\dfrac{{{\text{b}}\left( {\text{x}} \right){\text{ + c}}\left( {{\text{100 - x}}} \right)}}{{{\text{100}}}}$
Here, a represents the average atomic mass, b and c denotes the masses of the two isotopes and x denotes the percent abundance of the first isotope.
According to the question, boron has two naturally occurring isotopes ${}_{\text{5}}{{\text{B}}^{{\text{11}}}}$ which has a mass of 11 amu and ${}_{\text{5}}{{\text{B}}^{{\text{10}}}}$ which has a mass of 10 amu.
Also given, the average atomic mass of boron is ${\text{10}}{\text{.81amu}}$ .
Substitute these values of boron in the above equation and we get
${\text{10}}{\text{.81 = }}\dfrac{{11 \times \left( {\text{x}} \right){\text{ + 10}} \times \left( {{\text{100 - x}}} \right)}}{{{\text{100}}}}$
Factor the equation by which we will have
$
  {\text{10}}{\text{.81}} \times {\text{100 = }}11{\text{x + 10}} \times 100 - 10 \times {\text{x}} \\
   \Rightarrow {\text{1081 = }}11{\text{x + }}1000 - 10 \times {\text{x}} \\
 $
Bring the x terms together on one side of the equation and the non – x terms together on the other side of the equation.
$1081 - 1000 = 11{\text{x}} - 10{\text{x}}$
Rearrange the equation to get
$
  11{\text{x}} - 10{\text{x = }}1081 - 1000 \\
   \Rightarrow {\text{x = }}81 \\
 $
This is the abundance percentage of the first isotope ${}_{\text{5}}{{\text{B}}^{{\text{11}}}}$ having mass 11 amu.
Therefore, the abundance percentage of the second isotope ${}_{\text{5}}{{\text{B}}^{{\text{10}}}}$ having mass 10 amu is $100 - {\text{x}}$ which on being substituted by the value of x gives $100 - 81 = 19$ percent.

So, the correct option is C.

Note: Isotopes have the same physical and chemical properties and the same atomic number, i.e., the same number of protons. Due to the difference in their number of neutrons, the isotopes differ in their atomic weights, density, atomic volume, melting point and boiling point.On the other hand, isobars are the atoms of different elements having the same mass number but different atomic numbers. For example, argon and calcium have the same mass number 40 but different atomic numbers 18 and 20 respectively.