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Question 11

The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 168 X and 188 X in the sample?

Answer

Given, the normal atomicr mass of an example of a component X is 16.2u.
Two isotopes of component =   _{8}^{16}\textrm{X} and  _{8}^{18}\textrm{X}
Presently, Let's percent of isotope  _{8}^{16}\textrm{X} be x and percent of  _{8}^{18}\textrm{X} be 100 - x
In this way, According to the inquiry,

Average Atomic Mass :

16.2=16\times\frac{x}{100}+18\times\frac{100-x}{100}

16.2=\frac{16x}{100}+18-\frac{18x}{100}

-1.8=-\frac{2x}{100}

2x=180

x=90

Subsequently the level of isotope _{8}^{16}\textrm{X}  is 90 % and the percentage isotope  _{8}^{16}\textrm{X} is 10%.

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