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Chapter 2: Problem 88

Atoms are spherical and so when silver atoms pack together to form silver metal, they cannot fill all the available space. In a sample of silver metal, approximately \(26.0 \%\) of the sample is empty space. Given that the density of silver metal is \(10.5 \mathrm{g} / \mathrm{cm}^{3}\), what is the radius of a silver atom? Express your answer in picometers.

Short Answer

Expert verified
The radius of a silver atom is approximately \(144.4 \mathrm{pm}\).

Step by step solution

01

Identify the used constants

The atomic weight of silver is \(107.87 \mathrm{~g/mole}\) and the Avogadro number is \(6.022 \times 10^{23} \mathrm{atoms/mole}\). The fraction of silver sphere occupied is \(74 \%\) which is \(0.74\) as a decimal.
02

Calculate the number of atoms

The number of atoms in \(1 \mathrm{cm}^{3}\) of silver can be calculated based on the fraction of space they occupy. This is done by the formula \(\text{{Number of atoms}} = \frac{{\text{{Density}} \times \text{{Avogadro's number}}}}{{\text{{Atomic weight}}}} = \frac{{10.5 \mathrm{g/cm^{3}} \times 6.022 \times 10^{23} \mathrm{atoms/mole}}}{{107.87 \mathrm{g/mole}}} \approx 5.855 \times 10^{22} \mathrm{atoms/cm^{3}}\).
03

Calculate the volume occupied by a single atom

The volume a single atom occupies is simply the total volume divided by the number of atoms. Since we know that the total volume for the silver atoms is \(74 \%\) of \(1 \mathrm{~cm^{3}}\), the volume of a single atom is \(\frac{{0.74 \mathrm{cm^{3}}}}{{5.855 \times 10^{22} \mathrm{atoms}}} \approx 1.263 \times 10^{-23} \mathrm{cm^{3/atom}}\).
04

Calculate the radius of the atom

Since the silver atom is spherical, the volume can be represented as \(\frac{4}{3}\pi r^{3}\). Solving this equation for \(r\) and substituting the previously calculated volume into this equation, we have \(r = \left(\frac{3V}{4\pi}\right)^{1/3} = \left(\frac{3 \times 1.263 \times 10^{-23} \mathrm{cm^{3/atom}}}{4\pi}\right)^{1/3} \approx 1.444 \times 10^{-8} \mathrm{cm}\). To convert centimeters to picometers, 1 centimeter is equal to \(1\times 10^{10}\) picometers. Hence, the radius of the silver atom is \(1.444 \times 10^{-8} \mathrm{cm} \times 1\times 10^{10} \mathrm{pm/cm} \approx144.4 \mathrm{pm}\).

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