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Chapter 10: Problem 67

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 390 ppm (parts per million, by volume; that is, \(390 \mathrm{~L}\) of every \(10^{6} \mathrm{~L}\) of the atmosphere are \(\left.\mathrm{CO}_{2}\right)\). What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?

Short Answer

Expert verified
The mole fraction of CO₂ in the atmosphere can be calculated using the simplified equation \(n=\frac{V}{V_{atm}}\). Plugging in the given values, we get: \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\). The mole fraction of CO₂ is equal to the ratio of moles of CO₂ to the total number of moles in the atmosphere: \(\chi_{CO₂} = \frac{n_{CO₂}}{n_{air} + n_{CO₂}}\). Since the sum of all mole fractions is 1, we can solve for \(\chi_{CO₂}\) by plugging in the values: \(\chi_{CO₂} = \frac{0.00039\,\text{L}}{1\,\text{L}}\). Thus, the mole fraction of CO₂ in the atmosphere is approximately \(3.9 \times 10^{-4}\).

Step by step solution

01

Convert ppm concentration to Liters

The given concentration of CO₂ is 390 ppm. In this case, ppm is by volume, so we need to convert it to liters. It means that for every 10^6 L of the atmosphere, there are 390 L of CO₂.
02

Calculate moles of CO₂

Now use the ideal gas law to calculate the moles of CO₂ from the given volume. The ideal gas law equation is: \(PV = nRT\) Where: - P = pressure of the gas - V = volume of the gas - n = moles of the gas - R = universal gas constant (\(8.314 \frac{J}{mol*K}\)) - T = temperature of the gas (in Kelvin) Considering that we only need the mole fraction, we can simplify. Since the pressure and temperature are the same in the atmosphere and CO₂, we can use a simplified equation: \(n=\frac{V}{V_{atm}}\) Where: - n = moles of CO₂ - V = volume of CO₂ - \(V_{atm}\) = total volume of the atmosphere Therefore, \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\)
03

Calculate the mole fraction of CO₂

Now that we have the moles of CO₂, we can find the mole fraction of CO₂. The mole fraction is the ratio of the number of moles of CO₂ to the total number of moles of the atmosphere. Assuming the atmosphere contains only CO₂ and air, the mole fraction is: \(\chi_{CO₂} = \frac{n_{CO₂}}{n_{air} + n_{CO₂}}\) We have calculated \(n = \frac{390\,\text{L}}{10^6\,\text{L}}\), and it is important to note that the sum of the mole fractions for all components in the atmosphere is equal to 1: \(1 = \frac{n_{CO₂}}{n_{air} + n_{CO₂}}\) We can now replace n with the value we found and solve for the mole fraction of CO₂: \(\chi_{CO₂} = \frac{\frac{390\,\text{L}}{10^6\,\text{L}}}{1}\)
04

Calculate the final mole fraction

Now simply plug the value of n into the equation and solve for \(\chi_{CO₂}\): \(\chi_{CO₂} = \frac{0.00039\,\text{L}}{1\,\text{L}}\) Therefore, the mole fraction of CO₂ in the atmosphere is \(3.9 \times 10^{-4}\) (approximately).

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Most popular questions from this chapter

An open-end manometer containing mercury is connected to a container of gas, as depicted in Sample Exercise \(10.2 .\) What is the pressure of the enclosed gas in torr in each of the following situations? (a) The mercury in the arm attached to the gas is \(15.4 \mathrm{~mm}\) higher than in the one open to the atmosphere; atmospheric pressure is 0.985 atm. (b) The mercury in the arm attached to the gas is \(12.3 \mathrm{~mm}\) lower than in the one open to the atmosphere; atmospheric pressure is \(0.99 \mathrm{~atm} .\)

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for Xe, calculate the radius of a Xe atom and compare it to the value found in Figure 7.6, \(1.30 \AA\) A. Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24{ }^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of \(Z\) n have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. (a) What happens to the volume of the container as a result of the reaction? Explain. (b) If the piston is not allowed to move, what happens to the pressure as a result of the reaction? [Sections 10.3 and 10.5 ]

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{Cl} .\) The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces \(83.16 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(73.30 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor at STP. A separate analysis shows that the sample also contains \(16.44 \mathrm{mg}\) of Cl. (a) Determine the percent composition of the substance. (b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?
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