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Chapter 2: Problem 61

How many \(^{204} \mathrm{Pb}\) atoms are present in a piece of lead weighing \(215 \mathrm{mg} ?\) The percent natural abundance of \(204 \mathrm{Pb}\) is \(1.4 \%\)

Short Answer

Expert verified
The final answer after combining all the results would be the number of \(^{204}\)Pb atoms in 215 mg of lead.

Step by step solution

01

Calculate the number of Pb atoms in 215 mg of lead

Firstly, we need to convert 215 mg to grams by dividing it by 1000. So, we get 0.215 g of lead. Next, we use Avogadro’s number (\(6.022 \times 10^{23}\) atoms/mol) and the atomic mass of lead (207.2 g/mol) in order to figure out the number of atoms. The formula used here is (mass of lead / atomic mass of lead) * Avogadro’s number = number of Pb atoms.
02

Calculate \(^{204}Pb\) atoms

We know that the percent natural abundance of \(^{204}Pb\) is 1.4%. Hence, given so, the count of \(^{204}Pb\) atoms would simply be 1.4/100 multiplied by the total number of Pb atoms.
03

Final answer

Combine the results from step 1 and step 2 to find the number of \(^{204}\)Pb atoms in 215 mg of lead.

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