A mixture consisting of \(7.0 \mathrm{~g}\) of \(\mathrm{CO}\) and \(10.0 \mathrm{~g}\) of \(\mathrm{SO}_{2},\) two atmospheric pollutants, has a pressure of 0.33 atm when placed in a sealed container. What is the partial pressure of CO?

Short Answer

Expert verified
The partial pressure of \(\text{CO}\) is \0.203\text{ atm}\.

Step by step solution

01

- Calculate moles of \(\text{CO}\)

Use the molar mass of \(\text{CO}\) to convert grams to moles. The molar mass of \(\text{CO}\) is 28.01 g/mol. \Number of moles of \(\text{CO} = \frac{7.0 \text{ g}}{28.01 \text{ g/mol}} = 0.25 \text{ mol} \).
02

- Calculate moles of \(\text{SO}_{2}\)

Similarly, use the molar mass of \(\text{SO}_{2}\) to convert grams to moles. The molar mass of \(\text{SO}_{2}\) is 64.07 g/mol. \Number of moles of \(\text{SO}_{2} = \frac{10.0 \text{ g}}{64.07 \text{ g/mol}} = 0.156 \text{ mol} \).
03

- Total moles in the mixture

Add the moles of \(\text{CO}\) and \(\text{SO}_{2}\) to find the total moles in the mixture. \Total moles \(\text{n}_{\text{total}} = 0.25 \text{ mol} + 0.156 \text{ mol} = 0.406 \text{ mol} \).
04

- Calculate mole fraction of \(\text{CO}\)

The mole fraction \(\text{X}_{\text{CO}}\) is the ratio of moles of \(\text{CO}\) to the total moles. \Mole fraction \(\text{X}_{\text{CO}} = \frac{0.25 \text{ mol}}{0.406 \text{ mol}} = 0.616\).
05

- Calculate partial pressure of \(\text{CO}\)

The partial pressure of \(\text{CO}\) is found by multiplying its mole fraction by the total pressure. \Partial pressure of \(\text{CO} = \text{X}_{\text{CO}} \times P_{\text{total}} = 0.616 \times 0.33 \text{ atm} = 0.203 \text{ atm} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass
Molar mass is an important concept in chemistry, particularly when dealing with gas calculations like partial pressure. It represents the weight of one mole of a given substance. The unit for molar mass is grams per mole (g/mol). To find the molar mass, you sum the atomic masses of all the atoms in a molecule.
Molar mass gives you a way to convert between grams and moles. For example, the molar mass of carbon monoxide (CO) is calculated by adding the atomic masses of carbon (12.01 g/mol) and oxygen (16.00 g/mol) to get 28.01 g/mol.
Knowing how to calculate and use molar mass is crucial for converting grams to moles, which is needed for calculating the number of molecules involved in a reaction or process. This understanding sets the foundation for further calculations involving mole fraction and partial pressure.
Mole Fraction
Mole fraction is a way of expressing the concentration of a component in a mixture. It is the ratio of the number of moles of one component to the total number of moles in the mixture.
To find the mole fraction \(X_{\text{CO}}\) of carbon monoxide (CO) in a mixture, follow this formula:
\[ X_{\text{CO}} = \frac{\text{moles of CO}}{\text{total moles}} \]
In our example, if you have 0.25 moles of CO and the total moles in the mixture is 0.406, then the mole fraction is:
\[ X_{\text{CO}} = \frac{0.25}{0.406} = 0.616 \]
Mole fraction is a dimensionless quantity and is essential in calculating partial pressures of gases in a mixture.
Partial Pressure
Partial pressure describes the pressure exerted by a single type of gas in a mixture of gases. According to Dalton's Law, the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual gas.
To calculate the partial pressure of carbon monoxide (CO) in our example, we use its mole fraction and the total pressure of the gas mixture. The formula is:
\[ P_{\text{CO}} = X_{\text{CO}} \times P_{\text{total}} \]
Given that the mole fraction of CO is 0.616 and the total pressure is 0.33 atm, the partial pressure of CO becomes:
\[ P_{\text{CO}} = 0.616 \times 0.33 \text{ atm} = 0.203 \text{ atm} \]
Partial pressures are useful in many chemical processes and reactions, particularly those involving gases.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of Freon- \(12\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right.\) ) occupies \(25.5 \mathrm{~L}\) at \(298 \mathrm{~K}\) and \(153.3 \mathrm{kPa}\). Find its volume at \(\mathrm{STP}\).

Combustible vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume \(\%\) of vapor that gives a combustible mixture is called the lower flammable limit (LFL). Generally, the LFL is about half the stoichiometric mixture, which is the concentration required for complete combustion of the vapor in air. (a) If oxygen is 20.9 vol \% of air, estimate the LFL for \(n\) -hexane, \(\mathrm{C}_{6} \mathrm{H}_{14}\). (b) What volume (in \(\mathrm{mL}\) ) of \(n\) -hexane \(\left(d=0.660 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is required to produce a flammable mixture of hexane in \(1.000 \mathrm{~m}^{3}\) of air at \(\mathrm{STP} ?\)

Naturally occurring uranium ore is \(0.7 \%\) by mass fissionable \({ }^{235} \mathrm{U}\) and \(99.3 \%\) by mass nonfissionable \({ }^{238} \mathrm{U}\). For its use as nuclear reactor fuel, the amount of \({ }^{235} \mathrm{U}\) must be increased relative to the amount of \({ }^{238} \mathrm{U}\). Uranium ore is treated with fluorine to yield a gaseous mixture of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) that is pumped through a series of chambers separated by porous barriers; the lighter \({ }^{235} \mathrm{UF}_{6}\) molecules \((\mathscr{A}=349.03 \mathrm{~g} / \mathrm{mol}\) ) effuse through each barrier faster than molecules of \({ }^{238} \mathrm{UF}_{6}(\mathscr{A}=352.04 \mathrm{~g} / \mathrm{mol}),\) until the final mixture obtained is \(3-5 \%\) by mass \({ }^{235} \mathrm{UF}_{6} .\) This process generated \(33 \%\) of the world's enriched uranium in 2008 but has now been replaced with a less expensive centrifuge process. Calculate the ratio of the effusion rates of \({ }^{235} \mathrm{UF}_{6}\) to \({ }^{238} \mathrm{UF}_{6}\).

When a car accelerates quickly, the passengers feel a force that presses them back into their seats, but a balloon filled with helium floats forward. Why?

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure is tripled (at constant \(T\) ). (b) The absolute temperature is increased by a factor of 3.0 (at constant \(P\) ). (c) Three more moles of the gas is added (at constant \(P\) and \(T\) ).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free