Dalton's law states that the total pressure, Pal of a mixture of gases in a container equals the sum of the pressures of each individual gas PR+++... The partial pressure of the first component. P, is equal to the mole fraction of this component, X₁, times the total pressure of the mixture R-X₁ x P The mole fraction, X. represents the concentration of the component in the gas mixture, so moles of component X₁n misture Part A Three gases (8.00 g of methane, CHI,, 18.0 g of ethane, C₂lle, and an unknown amount of propane. C₂H₂) were added to the same 10.0L container. At 23.0°C, the total pressure in the container is 4.20 atm. Calculate the partial pressure of each gas in the container Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. View Available Hings) PPP 2.982 Submit Part B Previous Answers Pe 2.27 10¹ A gaseous mixture of Oy and Ny contains 39.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 845 mmHg? Express you answer numerically in millimeters of mercury. View Available in Submit Previous Answers * Incorrect; Try Again AZ? < Return to Assignment Provide Feedback O? atm mmHg Review Constants I Periodic

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Chapter1: Chemical Foundations
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I need help with my chemistry question part A and B

 

**Dalton's Law of Partial Pressure**

Dalton’s law states that the total pressure, \( P_{\text{total}} \), of a mixture of gases in a container equals the sum of the pressures of each individual gas:

\[ 
P_{\text{total}} = P_1 + P_2 + P_3 + \ldots 
\]

The partial pressure of the first component, \( P_1 \), is equal to the mole fraction of this component, \( X_1 \), times the total pressure of the mixture:

\[ 
P_1 = X_1 \times P_{\text{total}} 
\]

The mole fraction, \( X \), represents the concentration of the component in the gas mixture, so:

\[ 
X_1 = \frac{\text{moles of component 1}}{\text{total moles in mixture}} 
\]

---

**Part A**

Three gases (80.0 g of methane, \( \text{CH}_4 \), 18.0 g of ethane, \( \text{C}_2\text{H}_6 \), and an unknown amount of propane, \( \text{C}_3\text{H}_8 \)) were added to the same 10.0 L container. At 23.0°C, the total pressure in the container is 4.20 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.

Input box for pressure values: `2.982 atm`

*Note: User input is shown as 2.982 atm.*

---

**Part B**

A gaseous mixture of \( \text{O}_2 \) and \( \text{N}_2 \) contains 30.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 845 mmHg? Express your answer numerically in millimeters of mercury.

Input box for partial pressure of oxygen: `2.27 \times 10^3 mmHg`

Response: "Incorrect; Try Again"

*Note: User had an incorrect attempt with input `2.27 \times 10^3 mmHg`.*

--- 

This exercise focuses on applying Dalton's Law to calculate the partial pressures of gases in a mixture.
Transcribed Image Text:**Dalton's Law of Partial Pressure** Dalton’s law states that the total pressure, \( P_{\text{total}} \), of a mixture of gases in a container equals the sum of the pressures of each individual gas: \[ P_{\text{total}} = P_1 + P_2 + P_3 + \ldots \] The partial pressure of the first component, \( P_1 \), is equal to the mole fraction of this component, \( X_1 \), times the total pressure of the mixture: \[ P_1 = X_1 \times P_{\text{total}} \] The mole fraction, \( X \), represents the concentration of the component in the gas mixture, so: \[ X_1 = \frac{\text{moles of component 1}}{\text{total moles in mixture}} \] --- **Part A** Three gases (80.0 g of methane, \( \text{CH}_4 \), 18.0 g of ethane, \( \text{C}_2\text{H}_6 \), and an unknown amount of propane, \( \text{C}_3\text{H}_8 \)) were added to the same 10.0 L container. At 23.0°C, the total pressure in the container is 4.20 atm. Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane. Input box for pressure values: `2.982 atm` *Note: User input is shown as 2.982 atm.* --- **Part B** A gaseous mixture of \( \text{O}_2 \) and \( \text{N}_2 \) contains 30.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 845 mmHg? Express your answer numerically in millimeters of mercury. Input box for partial pressure of oxygen: `2.27 \times 10^3 mmHg` Response: "Incorrect; Try Again" *Note: User had an incorrect attempt with input `2.27 \times 10^3 mmHg`.* --- This exercise focuses on applying Dalton's Law to calculate the partial pressures of gases in a mixture.
Expert Solution
Step 1

According to Dalton's law of partial pressures,

the total pressure by a mixture of gases is equal to the sum of the partial pressures of each of the constituent gases.

The partial pressure is defined as the pressure each gas would exert if it alone occupied the volume of the mixture at the same temperature.

 

Partial pressure of a gas = Mole fraction of that gas × Total Pressure.

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