Question

Naturally occurring B consists of two isotopes, whose atomic weights are 10.01 and 11.01. The atomic weight of natural boron is 10.81. The % of each isotope in natural boron is:
(a) % of isotope of mass 10.01 = 20, % of isotope of mass 11.01 = 80
(b) % of isotope of mass 10.01 = 30, % of isotope of mass 11.01 = 70
(c) % isotope of mass of 10.01 = 50, % of isotope of mass 11.01 = 50
(d) None of these.

Answer 1

Hint: We know that the atomic weight of substance is given in relative terms. It is the average weight of the atoms of the element relative to a carbon atom, taken as 12. But in the case of isotopes there might be different atomic masses.

Complete Step by step answer:
So, we have given an element B which has two isotopes. Let us understand what isotopes are. Isotopes are variants of a particular element in which they have the same number of protons but differ in the number of neutrons in the atom. So, here as we said we have isotopes which weigh 10.01 and 11.01.
From the given weight we can identify its relative atomic mass. Then the relative atomic mass depends upon the relative abundance of various isotopes of that particular element. Suppose an element consists of two isotopes and average atomic mass is equal to
$ = \dfrac{{Relative\;abundance\left( 1 \right) \times Atomic\;mass\left( 1 \right) + Relative\;abundance\left( 2 \right) \times Atomic\;mass\left( 2 \right)}}{{Relative\;abundance\left( 1 \right) + Relative\;abundance\left( 2 \right)}}$

So, we have to find the relative abundance of both isotopes of boron.
Therefore, let us consider the relative abundance of type 1 as $a$, then the relative abundance of type 2 will be $\left( {100 - a} \right)$ since they are in its percentage value.
Atomic mass of isotopes are given as
$Atomic\;weight\;\left( 1 \right) = 10.01$
$\Rightarrow Atomic\;weight\;\left( 2 \right) = 11.01$
Also, the atomic weight of boron is 10.81 which is given in the question.
Now, substituting this values to the equation of average atomic mass
$10.81 = \dfrac{{10.01 \times a + 11.01\left( {100 - a} \right)}}{{100}}$
Since we are taking the relative abundance in %, the total abundance will be 100%.
Solving the above equation
$1081 = 10.01a + 1101 - 11.01a$
$\Rightarrow a = 20$
That means the relative abundance of isotopes having atomic weight 10.01 is 20%.
Therefore, relative abundance of type 2 = $100 - 20 = 80\% $
The relative abundance of isotopes having atomic weight 11.01 is 80%.

Hence, option (a) is correct.

Note: Isotopes are also known as nuclides, most of the elements that are found in nature are combinations of several isotopes. We use different isotopes in fields like carbon dating, nuclear reactors and in medicine purposes.