Question

An element E has two isotopes E-1 and E-2 which have atomic masses 18 u and 16 u respectively. If relative abundances of E-1 and E-2 are in the ratio of 9 : 1, then the average atomic mass of E is

• A. 16.2 u
• B. 16.8 u
• C. 17.2 u
• D. 17.8 u

Answer 1

The correct option is D 17.8 u

Atomic mass of E-1 isotope = 18 u
Atomic mass of E-2 isotope = 16 u
The relative abundances of E-1 isotope and E-2 isotope are in the ratio of 9 : 1.
Let the abundance of E-1 isotope be x% and that of E-2 isotope be (100 – x)%.
Now, according to the question:
$\frac{x}{100-x}=\frac{9}{1}$
x = 900 – 9x
10x = 900
x = 90
Therefore, x is 90% and (100 – x) is 10%.
$\text{Average atomic mass of E}=\frac{\text{Atomic mass of E-1 x x\% + Atomic mass of E-2 x (100 - x)\%}}{100}$
$=\frac{18u\times 90+16u\times 10}{100}$
$=\frac{1620u+160u}{100}$
= 17.8 u
Hence, the correct answer is option (d).