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Chapter 2: Problem 60

How many atoms are present in a \(75.0 \mathrm{cm}^{3}\) sample of plumber's solder, a lead-tin alloy containing \(67 \% \mathrm{Pb}\) by mass and having a density of \(9.4 \mathrm{g} / \mathrm{cm}^{3} ?\)

Short Answer

Expert verified
The total number of atoms in the sample is approximately \(1.37 \times 10^{24}\) atoms .

Step by step solution

01

Calculate the total mass of the sample

Multiply the density by the volume, \(9.4 \mathrm{g} / \mathrm{cm}^{3} \times 75.0 \mathrm{cm}^{3} = 705 g\). The total mass of the sample is thus 705 g.
02

Determine the mass of element Pb(Lead)

Multiply the percentage of lead by the total mass, \(67 \% \times 705 g = 472.35\, g\). Thus the mass of Pb in the solder is 472.35 g.
03

Calculate the number of moles of Pb(Lead)

Divide the mass of lead by the molar mass of lead, \(472.35 g/207.2 g/mol = 2.2806 \, moles\). The number of moles of lead therefore is 2.2806 moles.
04

Determine the number of atoms of Pb(Lead)

Multiply the number of moles by Avogadro's number \(2.2806 moles \times 6.022 \times 10^{23} atoms/mole = 1.37256872 \times 10^{24}\, atoms\). Hence, the total number of atoms in the sample is \(1.37256872 \times 10^{24}\, atoms\).

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