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Chapter 10: Problem 122

Argon has a cubic closest packed structure as a solid. Assuming that argon has a radius of \(190 . \mathrm{pm}\), calculate the density of solid argon.

Short Answer

Expert verified
The density of solid argon is approximately 1.72 g/ cm³.

Step by step solution

01

Calculate the edge length of the unit cell

Since argon has a cubic closest packed (ccp) structure, its unit cell is face-centered cubic (fcc). In a face-centered cubic structure, the edge length 'a' and the radius 'r' are related by the formula: \(a = 2\sqrt{2}r\). The radius of argon is given as 190 pm. Plug in the value of r to find the edge length of the unit cell: \[a = 2\sqrt{2}(190\,\text{pm})\] \[a \approx 536.66\,\text{pm}\]
02

Calculate the volume of the unit cell

The volume V of a cubic unit cell can be calculated using the edge length 'a'. The formula for the volume is: \(V = a^3\). Plug in the value of a obtained in the previous step to calculate the volume of the unit cell: \[V = (536.66\,\text{pm})^3\] \[V \approx 1.54\times 10^8\,\text{pm}^3\]
03

Calculate mass of argon atoms in the unit cell

In an fcc unit cell, there are four atoms per unit cell. The molar mass of argon is 39.95 g/mol. We can calculate the mass of one argon atom by dividing the molar mass by Avogadro's number (\(6.022\times10^{23}\,\text{atoms/mol}\)): \[mass_{Ar} = \frac{39.95\,\text{g/mol}}{6.022\times10^{23}\,\text{atoms/mol}} \approx 6.64\times10^{-23}\,\text{g/atom}\] To find the mass of four argon atoms in the unit cell, multiply the mass of one argon atom by 4: \[mass_{unit\ cell} = 4 (6.64\times10^{-23}\,\text{g/atom}) = 2.656\times10^{-22}\,\text{g}\]
04

Convert the volume of the unit cell to cm³

To calculate the density, we need to convert the volume of the unit cell from \(\text{pm}^3\) to cm³. We can use the conversion factor of \(1\,\text{cm} = 10^{12}\,\text{pm}\): \[V\,(\text{cm}^3) = \frac{1.54\times10^8\,\text{pm}^3}{(10^{12}\,\text{pm/cm})^3} = 1.54\times10^{-16}\,\text{cm}^{3}\]
05

Calculate the density of solid argon

Finally, we can calculate the density of solid argon using the formula \(\rho = \frac{mass}{volume}\): \[\rho_{Ar} = \frac{2.656\times10^{-22}\,\text{g}}{1.54\times10^{-16}\,\text{cm}^3} \approx 1.72\,\text{g/cm}^3\] The density of solid argon is approximately 1.72 g/ cm³.

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Most popular questions from this chapter

A \(20.0-\mathrm{g}\) sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with \(100.0 \mathrm{~g}\) water at \(80.0^{\circ} \mathrm{C}\). Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are \(2.03\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

The \(\mathrm{CsCl}\) structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 67 ). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

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MnO has either the \(\mathrm{NaCl}\) type structure or the \(\mathrm{CsCl}\) type structure (see Exercise 67). The edge length of the \(\mathrm{MnO}\) unit cell is \(4.47 \times 10^{-8} \mathrm{~cm}\) and the density of \(\mathrm{MnO}\) is \(5.28 \mathrm{~g} / \mathrm{cm}^{3}\) a. Does \(\mathrm{Mn} \mathrm{O}\) crystallize in the \(\mathrm{NaCl}\) or the \(\mathrm{CsCl}\) type structure? b. Assuming that the ionic radius of oxygen is \(140 . \mathrm{pm}\), estimate the ionic radius of manganese.
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