Question

The relative abundance of two rubidium isotopes of atomic weights $85$ and $87$ are $75\% $ and $25\% $ respectively. The average atomic weight of rubidium is:
A: $75.5$
B: $85.5$
C: $86.5$
D: $87.5$

Answer 1

Hint: Atomic weight of an element is the weight which is calculated by adding the multiplication of relative abundance and atomic weights of various isotopes. Isotopes are the atoms of the same element which have the same atomic number and different mass number.

Formula used: Average atomic mass$ = \sum\nolimits_{i = 1}^{i = n} {\left( {mas{s_{\left( i \right)}} \times abundance{e_{\left( i \right)}}} \right)} $

Complete step by step answer:
Atomic mass is the mass of one mole atoms of a substance. There are $6.022 \times {10^{23}}$ atoms or molecules in one mole of a substance. This means the average atomic mass of rubidium is the mass of one mole isotopes of rubidium.
In this question we have given relative abundance and atomic weight of two isotopes of rubidium and we have to find the atomic weight of rubidium. The relative abundance of one isotope with atomic weight $85$ is $75\% $ and the atom with atomic weight $87$ is $25\% $. This means if we have one hundred moles, then $75$ of those one hundred moles have molecular weight $85$ and $25$ of those one hundred moles have molecular weight $87$ (as the abundance of the atom with atomic weight $85$ is $75\% $ and the atom with atomic weight $87$ is $25\% $). We can find the average atomic weight of rubidium with the following formula,
Average atomic mass$ = \sum\nolimits_{i = 1}^{i = n} {\left( {mas{s_{\left( i \right)}} \times abundance{e_{\left( i \right)}}} \right)} $
Atomic mass of first isotope$ = mas{s_{\left( 1 \right)}} = 85$
Atomic mass of second isotope$ = mas{s_{\left( 2 \right)}} = 87$
Relative abundance of first isotope$ = abundance{e_{\left( 1 \right)}} = 75$
Relative abundance of second isotope$ = abundance{e_{\left( 2 \right)}} = 25$
The above formula can be written as,
Average atomic mass$ = \left( {mas{s_{\left( 1 \right)}} \times abundance{e_{\left( 1 \right)}}} \right) + \left( {mas{s_{\left( 2 \right)}} \times abundance{e_{\left( 2 \right)}}} \right)$
(Expanding the summation symbol)
Now, substituting the values in this formula,
Average atomic mass$ = \left( {85 \times 75} \right) + \left( {87 \times 25} \right)$
Solving this we get,
Average atomic mass$ = 8550$
Since the relative abundance that we have taken is with respect to one hundred atoms (explained above). This means the mass that we have calculated is the mass of one hundred moles. Mass of one mole that is average atomic weight of rubidium will be equal to,
Average atomic weight of rubidium$ = \dfrac{{Average{\text{ }}atomic{\text{ }}mass}}{{100}}$
(We have divided it by hundred because the mass we have calculated is the mass of one hundred atoms)
So, Average atomic weight of rubidium$ = \dfrac{{8550}}{{100}} = 85.5$

So, the correct answer is Option B .

Note:
There is an Avogadro number of particles in one mole of a substance. Avogadro number is equal to $6.022 \times {10^{23}}$. Number of atoms or particles in one mole of a substance is fixed. This number doesn’t depend on the temperature and pressure conditions.