Ncert Exercises & Solutions

13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are \({ }_{12}^{24} \mathrm{Mg}\) (23.98504u), \({ }_{12}^{25} \mathrm{Mg}\) (24.98584u), and \({ }_{12}^{26} \mathrm{Mg}\) (25.98259u). The natural abundance of \({ }_{12}^{24} \mathrm{Mg}\) is 78.99% by mass. Calculate the abundances of the other two isotopes.

Hint: Average atomic mass is given by the weightage average of masses of isotopes.
Step 1: Find the weightage average of masses of isotopes.
Mass of magnesium \({ }_{12}^{24} \mathrm{Mg}\) isotope m₁= 23.98504 u
Mass of magnesium \({ }_{12}^{25} \mathrm{Mg}\) isotope m₂= 24.98584 u
Mass of magnesium

{}_{12}^{26}M g i s o t o p e, m_{3}

= 25.98259 u
Abundance of

{}_{12}^{24}M g, n_{1}

=78.99%

Let the abundance of

{}_{12}^{25}M g, n_{2}

=x%,
Therefore, the abundance of

{}_{12}^{26}M g, n_{3} = 100 - x - 78.99 % =

[21.01 - x] %

We have the relation for the average atomic mass as:

m = \frac{m_{1} n_{1} + m_{2} n_{2} + m_{3} n_{3}}{n_{1} + n_{2} + n_{3}}

24.312 = \frac{23.98504 \times 78.99 + 24.98584 \times . x + 25.98259 \times (21.01 - x)}{100}

2431.2 = 1894.5783096 + 24.98584 x + 545.8942159 - 25.98259 x

0.99675 x = 9.2725255

∴ x \approx 9.3 %

A n d 21.01 - x = 11.71 %

H e n c e, t h e a b u n d a n c e o f {}_{12}^{25}M g i s 9.3 % a n d t h a t o f {}_{12}^{26}M g i s 11.71 %

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions