A certain sample of element Z contains 60% of $69_{Z}$ and 40% of $71_{Z}$ . What is the relative atomic mass of element Z in this sample.

69.2

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69.8

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70.2

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Solution

The correct option is A 69.8
RAM= $\frac{60 \times 69 + 71 \times 40}{100}$

$= \frac{414 + 284}{10}$

$= \frac{698}{10} = 69.84$

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Similar questions

^{69} Z

^{71} Z

A certain sample of element Z contains 60% of $^{69} Z$ and 40% of $^{71} Z$ . What is the average atomic mass of element Z in this sample?

^{69} X

^{71} X

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