Natural copper has two isotopes, $^{63} Cu$ and $^{65} Cu$ (take their masses as 63 and 65), and an average atomic mass of 63.62. How many $^{65} Cu$ atoms are present in 12.724 g of natural copper? Take $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ . Give the answer in units of $1 0^{22}$ , to two decimal places.

Numerical answer type, answer in x10^22, tolerance 0.03.

Answer: 3.73 x10^22

Let the fraction of

X_{65} X_{2}^{265} Cu

f

63 (1 - f) + 65 f = 63.62

, so

f = 0.31

. Moles of copper atoms

= 12.724/63.62 = 0.200

. Atoms of

X_{65} X_{2}^{265} Cu

= 0.200 \times 0.31 \times 6.02 \times 1 0^{23} = 3.73 \times 1 0^{22}

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