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Chapter 10: Problem 50

(a) Calculate the density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.

Short Answer

Expert verified
(a) The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) is approximately \(4.00 \mathrm{~g/L}\). (b) The molar mass of the vapor with a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr is approximately \(96.07 \mathrm{~g/mol}\).

Step by step solution

01

Write down the Ideal Gas Law formula

The Ideal Gas Law formula is \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
02

Convert temperature to Kelvin

To convert the given temperature of \(21^{\circ} \mathrm{C}\) to Kelvin, use the formula: \(T(K) = T(^{\circ} \mathrm{C}) + 273.15\). Therefore, \(T = 21 + 273.15 = 294.15\mathrm{~K}\).
03

Convert pressure to atm

To convert the given pressure of 707 torr to atm, use the conversion factor 1 atm = 760 torr. Therefore, \(P = \frac{707}{760} \mathrm{~atm}\).
04

Rewrite the formula in terms of density

Density (\(\rho\)) is defined as \(\rho = \frac{m}{V}\), where m is the mass. We can also represent mass as the product of moles and molar mass (M): \(m = nM\). Substituting this expression in the density formula, we get: \(\rho = \frac{nM}{V}\). Now, rearrange the Ideal Gas Law formula to obtain \(\frac{n}{V} = \frac{P}{RT}\) and substitute this expression into the density formula: \(\rho = \frac{PM}{RT}\).
05

Substitute values and calculate density

Use the sulfur hexafluoride molar mass, M = 146.06 g/mol, and the given temperature and pressure values, \(T = 294.15\mathrm{~K}\) and \(P = \frac{707}{760} \mathrm{~atm}\), in the density formula: \(\rho = \frac{PM}{RT}\). We have: \(\rho = \frac{(\frac{707}{760})(146.06)}{(0.0821)(294.15)}\). On calculating, we get \(\rho \approx 4.00 \mathrm{~g/L}\). (a) The density of sulfur hexafluoride gas at 707 torr and \(21^{\circ} \mathrm{C}\) is approximately \(4.00 \mathrm{~g/L}\). (b) Calculate the molar mass of a vapor that has a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr.
06

Convert temperature to Kelvin

To convert the given temperature of \(12^{\circ} \mathrm{C}\) to Kelvin, use the formula: \(T(K) = T(^{\circ} \mathrm{C}) + 273.15\). Therefore, \(T = 12 + 273.15 = 285.15\mathrm{~K}\).
07

Convert pressure to atm

To convert the given pressure of 743 torr to atm, use the conversion factor 1 atm = 760 torr. Therefore, \(P = \frac{743}{760} \mathrm{~atm}\).
08

Rearrange the formula to find molar mass

Use the density formula from Step 4: \(\rho = \frac{PM}{RT}\). Rearrange it to find molar mass, M: \(M = \frac{\rho RT}{P}\).
09

Substitute values and calculate molar mass

Use the given density (\(\rho = 7.135 \mathrm{~g/L}\)), temperature (\(T = 285.15\mathrm{~K}\)), and pressure (\(P = \frac{743}{760} \mathrm{~atm}\)), in the molar mass formula: \(M = \frac{\rho RT}{P}\). We have: \(M = \frac{(7.135)(0.0821)(285.15)}{(\frac{743}{760})}\). On calculating, we get \(M \approx 96.07 \mathrm{~g/mol}\). (b) The molar mass of the vapor with a density of \(7.135 \mathrm{~g} / \mathrm{L}\) at \(12{ }^{\circ} \mathrm{C}\) and 743 torr is approximately \(96.07 \mathrm{~g/mol}\).

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Most popular questions from this chapter

A neon sign is made of glass tubing whose inside diameter is \(2.5 \mathrm{~cm}\) and whose length is \(5.5 \mathrm{~m}\). If the sign contains neon at a pressure of \(1.78\) torr at \(35^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h .\) )

Magnesium can be used as a "getter" in evacuated enclosures, to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.382 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

A sample of \(4.00 \mathrm{~mL}\) of diethylether \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\), density \(=0.7134 \mathrm{~g} / \mathrm{mL}\) ) is introduced into a 5.00-L vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(P_{N_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.208 \mathrm{~atm}\). The temperature is held at \(35.0^{\circ} \mathrm{C}\), and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) are oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane described above and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

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