Question

The density of liquid mercury at \(20^{\circ} \mathrm{C}\) is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\), its vapor pressure is \(1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}\). (a) What volume (in \(\mathrm{cm}^{3}\) ) is occupied by one mole of \(\mathrm{Hg}(l)\) at \(20^{\circ} \mathrm{C}\) ? (b) What volume (in \(\mathrm{cm}^{3}\) ) is occupied by one mole of \(\mathrm{Hg}(\mathrm{g})\) at \(20^{\circ} \mathrm{C}\) and the equilibrium vapor pressure? (c) The atomic radius of \(\mathrm{Hg}\) is \(0.155 \mathrm{~nm}\). Calculate the volume (in \(\mathrm{cm}^{3}\) ) of one mole of \(\mathrm{Hg}\) atoms \(\left(V=4 \pi r^{3} / 3\right)\). (d) From your answers to (a), (b), and (c), calculate the percentage of the total volume occupied by the atoms in \(\mathrm{Hg}(l)\) and \(\mathrm{Hg}(g)\) at \(20^{\circ} \mathrm{C}\) and \(1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}\)

Short answer

Answer: The atoms in liquid mercury occupied approximately \(63.7\%\) of the total volume, and the atoms in gaseous mercury occupied approximately \(6.52\times10^{-6}\%\) of the total volume at \(20^{\circ} \mathrm{C}\) and \(1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}\).

Step-by-step solution

Calculate the volume of one mole of liquid mercury

To calculate the volume of one mole of liquid mercury, use the density equation: Density = mass/volume Using the given density of liquid mercury (\(13.6\mathrm{~g}/\mathrm{cm}^{3}\)), we can calculate the volume occupied by one mole (\(200.59\mathrm{~g}\)) of Hg. Volume = mass/density Volume = \(\frac{200.59\mathrm{~g}}{13.6\mathrm{~g}/\mathrm{cm}^{3}}\) Volume \(= 14.75\mathrm{~cm}^{3}\)

Calculate the volume of one mole of gaseous mercury

To find the volume occupied by one mole of gaseous mercury at the given conditions, use the ideal gas law equation: PV = nRT Here, P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. We are given pressure (\(1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}\)) and temperature (\(20^{\circ} \mathrm{C}\) or \(293.15 \mathrm{~K}\)), and we are dealing with one mole of mercury. First, convert pressure from \(\mathrm{mm} \mathrm{Hg}\) to \(\mathrm{atm}\). \(\displaystyle\frac{1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}}{760\mathrm{~mm}\mathrm{Hg}/\mathrm{atm}} = 1.579 \times 10^{-6}\mathrm{~atm}\) Now, use the ideal gas law with R in units of \(\mathrm{L} \cdot \mathrm{atm}/\mathrm{mol} \cdot \mathrm{K}\): \(1.579 \times 10^{-6}\mathrm{~atm} \cdot V = 1 \cdot 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 293.15\mathrm{~K}\) Rearranging to solve for V: \(V=\displaystyle\frac{1 \cdot 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 293.15\mathrm{~K}}{1.579 \times 10^{-6}\mathrm{~atm}}\) \(V \approx 144000 \mathrm{L}\) Convert the volume to \(\mathrm{cm}^{3}\). \(144000 \mathrm{L} \cdot \displaystyle\frac{1000\mathrm{~cm}^{3}}{1\mathrm{L}} = 144000000\mathrm{~cm}^{3}\)

Calculate the volume of one mole of mercury atoms

To calculate the volume occupied by one mole of mercury atoms, use the given atomic radius and the formula for the volume of a sphere: Volume of one atom = \(\displaystyle\frac{4\pi r^{3}}{3}\) Where r is the radius (in this case, \(0.155\mathrm{~nm}\), which is equal to \(1.55\times10^{-8}\mathrm{~cm}\)). Volume of one atom = \(\displaystyle\frac{4\pi (1.55\times10^{-8}\mathrm{~cm})^{3}}{3} = 1.56 \times 10^{-23}\mathrm{~cm}^{3}\) We are looking for the volume occupied by one mole of mercury atoms. Since there are \(6.022 \times 10^{23}\) atoms in a mole: Volume of one mole = \((1.56 \times 10^{-23}\mathrm{~cm}^{3}) \cdot (6.022 \times 10^{23})\) Volume of one mole \(= 9.39\mathrm{~cm}^{3}\)

Calculate the percentage for liquid and gaseous mercury

Calculate the percentage of the total volume occupied by the atoms in liquid and gaseous mercury. For liquid mercury: Percentage = \(\displaystyle\frac{9.39\mathrm{~cm}^{3}}{14.75\mathrm{~cm}^{3}} \times 100\% \approx 63.7\%\) For gaseous mercury: Percentage = \(\displaystyle\frac{9.39\mathrm{~cm}^{3}}{144000000\mathrm{~cm}^{3}} \times 100\% \approx 6.52 \times 10^{-6}\%\) The atoms in liquid mercury occupied approximately \(63.7\%\) of the total volume, and the atoms in gaseous mercury occupied approximately \(6.52\times10^{-6}\%\) of the total volume at \(20^{\circ} \mathrm{C}\) and \(1.2 \times 10^{-3} \mathrm{~mm} \mathrm{Hg}\).

Key concepts

Molar Volume

Molar volume is a concept in chemistry that represents the volume one mole of a substance occupies. It is particularly important when dealing with gases, as it is standard to compare their volume at a common temperature and pressure. For instance, at standard temperature and pressure (STP), which is defined as 0 degrees Celsius and 1 atmosphere of pressure, one mole of any gas occupies 22.4 liters.

The problem provided allows us to explore the molar volume in the context of liquid mercury (Hg), which is known to have a much higher density than many other substances, especially gases. By establishing that the volume occupied by one mole of Hg at a certain temperature and pressure is 14.75 \(\text{cm}^3\), we have the molar volume for mercury in its liquid state at the conditions stated.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. It is usually expressed as PV = nRT, where P represents pressure, V represents volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

For the exercise involving gaseous mercury, the ideal gas law helps calculate the volume occupied by one mole of mercury gas given the temperature and pressure. This relationship assumes that the gas behaves ideally, which means that the particles do not interact with each other and occupy no volume themselves, an assumption that is more accurate for gases at low pressure and high temperature.

Density of Substances

Density is a measure of how much mass is contained in a given volume of a substance and is commonly expressed in terms of grams per cubic centimeter (\(\text{g/cm}^3\)) for solids and liquids, and grams per liter (\(\text{g/L}\)) for gases. In the context of the problem, knowing the density of mercury allows us to calculate the volume occupied by a specific mass of the substance. It is a critical concept when comparing substances or converting between mass and volume.

Density is inversely proportional to volume given a constant mass. This means that as the density of a substance increases, the volume that a certain mass of the substance occupies decreases. The density of liquid mercury is exceptionally high, which explains why a mole of mercury occupies a relatively small volume.

Atomic Radius

Atomic radius refers to the distance from the center of an atom to the outermost electrons. It is a periodic property that can give insights into many aspects of an element's chemistry, such as its reactivity and how it bonds with other atoms. In metallic elements like mercury, the atomic radius can also indicate how closely packed the atoms are in its solid form.

In our exercise, using the atomic radius of mercury, we calculated the volume of a single Hg atom and then extrapolated this to find the volume occupied by one mole of Hg atoms. This illustrates the concept of atomic packing and can be related to the density of the substance in its solid or liquid form.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system. It indicates a liquid's tendency to evaporate, and high vapor pressure implies high volatility. The given vapor pressure for mercury at 20 degrees Celsius is low, reflecting mercury's relatively low volatility.

In practical terms, when conducting the exercise's calculations for gaseous mercury, we convert the vapor pressure from mmHg to atmospheres to use in the ideal gas law formula. Vapor pressure is key to understanding a substance's physical properties and provides insights into phenomena like boiling, melting points, and phase transitions.