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Question: Average atomic mass of Mg is approximately: ![](https://www.vedantu.com/question-sets/aae9cc38-9be...

Average atomic mass of Mg is approximately:

(A) 25.0
(B) 24.5
(C) 25.2
(D) 25.8

Explanation

Solution

The graph gives us abundances of different isotopes of magnesium. The average atomic weight of any element is the sum of the masses of the isotopes multiplied with their respective percentage abundances.

Formula used:
-Average atomic weight:
= i=1n(mass(i))(abundance(i))\sum\limits_{i = 1}^n {(mas{s_{(i)}})(abundanc{e_{(i)}})} (1)
Where, mass = atomic weight of the isotope
Abundance = percentage abundance in which the respective isotope is present.

Complete answer:
-Isotopes are those atoms of a chemical element which have the same atomic number, same position in the periodic table and nearly the same chemical properties but they differ in the mass number and physical properties.
Isotopes of an element have the same number of electrons and protons but differ in the number of neutrons.
Example: -Carbon has 3 isotopes: C-12, C-13, C-14
-Hydrogen has 3 isotopes: H-1, D-2, T-3
-Bromine has 2 stable isotopes: Br-79, Br-81
-Magnesium has 3 isotopes: Mg-24, Mg-25, Mg-26
-From the graph given in the question we can tell the relative abundances of each isotope. They will be: (a) Mg-24: 70%
(b) Mg-25: 10%
(c) Mg-26: 20%
We use equation (1) to calculate the average atomic mass of Mg atom:
Average atomic weight=i=1n(mass(i))(abundance(i))\sum\limits_{i = 1}^n {(mas{s_{(i)}})(abundanc{e_{(i)}})}
= \left\\{ {24 \times \frac{{70}}{{100}}} \right\\} + \left\\{ {25 \times \frac{{10}}{{100}}} \right\\} + \left\\{ {26 \times \frac{{20}}{{100}}} \right\\}
= 16.8 + 2.5 + 5.2
= 24.5
Average atomic weight of Mg is 24.5 amu.

So, the correct option will be: (B) 24.5

Note:
Isotopes are those which have the same atomic number but different mass number. Isobars are those atoms which have the same mass number but different atomic number. Isotones are those which have the same neutron number but different proton number. Don’t confuse them with each other.