Question

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?

• A. 69.2
• B. 69.8
• C. 70
• D. 70.2

Answer 1

The correct option is B 69.8

The mass of the first isotope( $m_1$) = 69

The percentage of the first isotope( $p_1$) = 60

The mass of the second isotope ($m_2$) = 71

The percentage of the second isotope ($p_2$) = 40

Then, the average atomic mass of the element

$=\frac{m_1\times p_1+m_2\times p_2}{100}$

$=\frac{60\times 69+71\times 40}{100}=69.8$