Problem 60 - Chemistry: The Central Science

Question

Breathing air that contains \(4.0 \%\) by volume \(\mathrm{CO}_{2}\) over time causes rapid breathing, throbbing headache, and nausea, among other symptoms. What is the concentration of \(\mathrm{CO}_{2}\) in such air in terms of (a) mol percentage, (b) molarity, assuming \(1 \mathrm{~atm}\) pressure and a body temperature of \(37^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

The concentration of CO₂ in such air is (a) 4.0% mol percentage, and (b) approximately 0.00158 mol/L, assuming 1 atm pressure and a body temperature of 37°C.

Step 1: Understanding Volume Percent to Mol Percent

The given composition is in volume percent, which means the mol percentages will be the same since all the components of the mixture are ideal gases and we have equal volume ratios. Therefore, the mol percentage of CO₂ is also \(4.0\%\). ## Part (b): Molarity of CO₂ ##

Step 2: Understanding Ideal Gas Law and known parameters

Recall that the Ideal Gas Law is given by the equation: \(PV = nRT\) where 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 the temperature as \(37^{\circ}C\) and pressure as \(1\ atm\). We will first need to convert the temperature to Kelvin.

Step 3: Convert temperature to Kelvin

To convert the temperature from Celsius to Kelvin, add 273.15: \(T(K) = T(^{\circ}C) + 273.15\) \(T = 37^{\circ}C + 273.15\) \(T = 310.15 K\) Now, we can use the Ideal Gas Law to find the molarity.

Step 4: Find molarity using Ideal Gas Law and mole percentage

From the mole percentage of \(4.0 \%\) that we calculated in Part (a), we know that in a 1 mole sample of air, there will be \(0.04\) moles of CO₂. Let's first find the total number of moles for 1 mole of air at the given temperature and pressure (from the Ideal Gas Law): \(PV = nRT\) \(n = \frac{PV}{RT}\) We use pressure in atmosphere, so let's first convert the Ideal Gas Constant to the appropriate units: \(R = 0.0821 \frac{L \cdot atm}{mol \cdot K}\) Now, plug in the values: \(n = \frac{1 \text{ atm} \cdot V}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 310.15 K}\) Solving for V gives: \(V = 25.39 L\) Now, let's find the molarity of CO₂ (moles per liter): Since in the 1 mole of air, we have \(0.04\) moles of CO₂, and we found that 1 mole of air at the given temperature and pressure occupies 25.39 L, we can find the molarity of CO₂: \[Molarity=\frac{n_{\mathrm{CO}_{2}}}{V}\] \[Molarity= \frac{0.04\ mol}{25.39\ L}\] \[Molarity=0.00158\ \frac{mol}{L}\] So, the concentration of CO₂ in terms of molarity is approximately \(0.00158\ \frac{mol}{L}\).

Key Concepts

Molarity

Molarity is a concept that refers to the concentration of a solute within a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. The formula to calculate molarity (\[ M \] , in moles per liter) is:
\[ M = \frac{n}{V} \]where:

• \( n \) is the number of moles of the solute
• \( V \) is the volume of the solution in liters

Molarity is useful because it provides a clear way to express the concentration of a solution regardless of the total volume. It is especially helpful in reactions involving solutions as it allows for precise calculations of needed quantities.
In this exercise, we used the molar percentage to determine the number of moles of carbon dioxide in 1 mole of air, calculated its volume using the Ideal Gas Law, and then determined the molarity of the CO₂ in that volume.

Mole Percentage

The mole percentage is a way of representing concentrations used especially in cases involving mixtures or solutions. It indicates the fraction of the total number of moles that one specific substance contributes to the entire mixture, expressed as a percentage. It is calculated by the following formula:
\[\text{Mole percentage} = \left( \frac{n_{component}}{n_{total}} \right) \times 100 \%\]where:

• \( n_{component} \) is the number of moles of the substance of interest
• \( n_{total} \) is the total number of moles in the mixture

In ideal gas mixtures, the mole percentage is directly related to the volume percentage due to the behavior of ideal gases. For this problem, the air containing CO₂ at a 4.0% volume percentage directly translates into a 4.0% mole percentage. This is because equal volumes of ideal gases, under the same conditions of temperature and pressure, contain equal numbers of moles.

Volume Percentage

Volume percentage is a measure of concentration that indicates the proportion of a particular gas among all the gas components within a mixture. In terms of ideal gases, the volume percentage is identical to the mole percentage. This equivalency arises from Avogadro’s law, which states that equal volumes of gases, at the same temperature and pressure, have the same number of molecules.
Volume percentage is represented by the fraction of the partial volume of the gas divided by the total volume of all gases present, multiplied by 100. The formula is:
\[\text{Volume percentage} = \left( \frac{V_{component}}{V_{total}} \right) \times 100 \%\]where:

• \( V_{component} \) is the volume of the gas in question
• \( V_{total} \) is the total volume of the gas mixture

In the given problem: the CO₂'s volume percentage was 4.0%, indicating that CO₂ makes up 4% of the total volume of air. Without deviation from ideal behavior, this is straightforward to interpret as the mole percentage, which complements our analysis of molarity of CO₂ in air.