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Chapter 5: Problem 51

A sample of oxygen is collected over water at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{~mm} \mathrm{Hg}\) in a 125-mL flask. The vapor pressure of water at \(22^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm} \mathrm{Hg}\). (a) What is the partial pressure of oxygen? (b) How many moles of dry gas are collected? (c) How many moles of wet gas are in the flask? (d) If \(0.0250 \mathrm{~g}\) of \(\mathrm{N}_{2}(\mathrm{~g})\) are added to the flask at the same temperature, what is the partial pressure of nitrogen in the flask? (e) What is the total pressure in the flask after nitrogen is added?

Short Answer

Expert verified
Answer: The total pressure in the flask after nitrogen is added is 1.16 atm (rounded to 2 decimal places).

Step by step solution

01

Apply Dalton's Law of Partial Pressures

Total Pressure in the flask, P_total = Pressure of oxygen (O2) + Vapor pressure of water (H2O) We need to find the Pressure of oxygen (O2). We do this by rearranging the formula: Pressure of oxygen (O2) = P_total - Vapor pressure of water (H2O)
02

Use the given values

We are given the total pressure as 752 mm Hg, and the vapor pressure of water at 22°C as 19.8 mm Hg. Replace them in the formula: Pressure of oxygen (O2) = 752 mm Hg - 19.8 mm Hg
03

Calculate the partial pressure of oxygen

Pressure of oxygen (O2)= 732.2 mm Hg So, the partial pressure of oxygen is 732.2 mm Hg. #b) Find the moles of dry gas collected#
04

Use the Ideal Gas Law

The Ideal Gas Law can be written as PV=nRT, where P is pressure, V is volume, n is moles of gas, R is the gas constant, and T is temperature in Kelvin. We will use this formula to find the moles of dry gas (oxygen). First convert the temperature from Celsius to Kelvin: T(K) = 22°C + 273.15 = 295.15 K Next, convert pressure of oxygen into atm, as the gas constant R value we will be using has units of atm L/(mol K). Pressure of oxygen (O2) = 732.2 mm Hg × (1 atm/760 mm Hg) = 0.9634 atm Now we have P = 0.9634 atm, V = 125 mL, T = 295.15 K, R = 0.0821 L atm/(mol K). We need to convert the volume from mL to L: V = 125 mL × (1 L/1000 mL) = 0.125 L. We will now solve for n (moles of O2):
05

Rearrange the Ideal Gas Law

n(O2) = PV/(RT) Substitute the values: n(O2) = (0.9634 atm × 0.125 L) / (0.0821 L atm/(mol K) × 295.15 K)
06

Calculate the moles of dry gas (oxygen)

n(O2) = 0.00515 mol Hence, there are 0.00515 moles of dry gas (oxygen) in the flask. #c) Find the moles of wet gas in the flask#
07

Find moles of water vapor

We need to use the Ideal Gas Law to find the moles of water vapor. We have already found the temperature (295.15 K) and R value (0.0821 L atm/(mol K)). The volume of water vapor is equal to the volume of the flask, 0.125 L. The pressure of water vapor equals the vapor pressure of water at 22°C which is 19.8 mm Hg. Convert pressure to atm: Pressure of water vapor (H2O) = 19.8 mm Hg × (1 atm/760 mm Hg)= 0.02605 atm Now, using the Ideal Gas Law: n(H2O) = PV/(RT) n(H2O) = (0.02605 atm × 0.125 L) / (0.0821 L atm/(mol K) × 295.15 K)
08

Calculate moles of water vapor

n(H2O) = 0.000139 mol
09

Calculate the total moles of wet gas in the flask

To find the total moles of wet gas (oxygen + water vapor), add the moles of oxygen (O2) and the moles of water vapor (H2O): n(wet gas) = n(O2) + n(H2O) = 0.00515 mol + 0.000139 mol = 0.005289 mol So, there are 0.00529 moles of wet gas in the flask. #d) Find the partial pressure of nitrogen in the flask (after nitrogen is added)#
10

Use the Ideal Gas Law

We will use the Ideal Gas Law again by solving for the partial pressure of nitrogen (N2) after it is added to the flask. We are given that 0.0250 g of nitrogen (N2) is added to the flask. First, convert the mass of nitrogen (N2) into moles using its molar mass (28.02 g/mol): n(N2) = 0.0250 g / 28.02 g/mol = 0.000892 mol So, we have 0.000892 moles of nitrogen (N2) added to the flask. The temperature and volume of nitrogen (N2) will be same as that of oxygen (O2).
11

Rearrange the Ideal Gas Law

P(N2) = n(N2)RT/V Substitute the values: P(N2) = 0.000892 mol × 0.0821 L atm/(mol K) × 295.15 K / 0.125 L
12

Calculate the partial pressure of nitrogen after it is added

P(N2) = 0.1754 atm So, the partial pressure of nitrogen after it is added to the flask is 0.1754 atm. #e) Find the total pressure in the flask after nitrogen is added#
13

Apply Dalton's Law of Partial Pressures

We now have the partial pressures of oxygen, water vapor, and nitrogen. To find the total pressure in the flask, we will apply Dalton's Law of Partial Pressures: P_total = P(O2) + P(H2O) + P(N2) Substitute the values (O2 and H2O pressures were calculated in earlier steps, and N2 pressure was obtained in the step d): P_total = 0.9634 atm + 0.02605 atm + 0.1754 atm
14

Calculate the total pressure in the flask

P_total = 1.16485 atm So, the total pressure in the flask after nitrogen is added is 1.16 atm (rounded to 2 decimal places).

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