Question 1

What is the pressure in a 1.0 L container into which has been placed 100 mL of a sample at 2.0 atm and 400 mL of a sample at 10 atm?

Accepted Answer

The total pressure is calculated using the formula for partial pressures: P_total = (P1V1 + P2V2) / V_total. Here, P1 = 2.0 atm, V1 = 0.1 L, P2 = 10 atm, V2 = 0.4 L, and V_total = 1.0 L. So, P_total = (2.0*0.1 + 10*0.4) / 1.0 = 4.2 atm.

Question 2

What is the partial pressure of Helium gas in a mixture of gases that contains 0.50 moles of helium, 1.5 moles of Argon, and 2.5 moles of Neon?
The total pressure of the mixture is 1.6 atm.

Accepted Answer

1. Find the total moles of the gas mixture:

0.50 mol He + 1.5 mol Ar + 2.5 mol Ne = 4.5 mol total

2. Find the mole fraction of Helium:

moles of Helium/total moles = 0.50 mol/4.5 mol = 0.111

3. Use the mole fraction of Helium and the total pressure to calculate the partial pressure of Helium:

partial pressure of Helium = mole fraction of Helium x total pressure

partial pressure of Helium = 0.111 x 1.6 atm

partial pressure of Helium = 0.178 atm, which is closest to choice B (0.18 atm).

Question 3

How many moles of argon gas are in a sample that is collected over water at 27 ° C (vapor pressure of water is 27 torr) if the volume of the gases is 430 mL and the atmospheric pressure is 754 torr?

Accepted Answer

The answer of How many moles of argon gas are...

Question 4

What is the partial pressure of He in a container that is at a total pressure of 2.33 atm and contains 1.3 g of Ar and 3.3 g of He?

Accepted Answer

The answer of What is the partial pressure of He...

Question 5

Exhibit 6-1 Consider a mixture of gases that includes 0.50 g of He, 5.0 g of Ne and 20 g of Ar contained in a 5.0 L vessel at 25 ° C to answer the following question(s). Molar Mass(He) = 4.0 g/mol, Molar Mass(Ne) = 20 g/mol, and Molar Mass(Ar) = 40 g/mol
-Refer to Exhibit 6-1. What is the mole fraction of Ne in this mixture?

Accepted Answer

The answer of Exhibit 6-1 Consider a mixture of gases...

Question 6

The total pressure exerted by a mixture of the three gases, oxygen, nitrogen, and water vapor, is 742 mm of Hg. The partial pressure of oxygen and nitrogen in the sample are 581 mm of Hg and 143 mm of Hg respectively. What is the partial pressure of water vapor in this mixture?

Accepted Answer

The total pressure of a gas mixture is equal to the sum of the partial pressures of its components. Therefore, the partial pressure of water vapor can be calculated as 742 mm Hg (total pressure) - 581 mm Hg (oxygen) - 143 mm Hg (nitrogen) = 18 mm Hg.

Question 7

What number of moles of hydrogen gas are in a 330 mL sample collected over water if the temperature is 27 ° C (vapor pressure of water at 27 ° C is 27 torr) and the atmospheric pressure is 755 torr?

Accepted Answer

The answer of What number of moles of hydrogen gas...

Question 8

A chemical engineer places a mixture of noble gases consisting of 10.0 grams of helium and 20.0 grams of Neon in a piston cylinder assembly. What is the mole fraction of Neon in this mixture?

Accepted Answer

The mole fraction of Neon is calculated by dividing the moles of Neon by the total moles of the mixture. The molar mass of Helium is approximately 4 g/mol, and for Neon, it is approximately 20 g/mol. Thus, there are 10.0 g / 4 g/mol = 2.5 moles of Helium and 20.0 g / 20 g/mol = 1.0 mole of Neon. The total moles in the mixture = 2.5 moles (He) + 1.0 mole (Ne) = 3.5 moles. The mole fraction of Neon = moles of Neon / total moles = 1.0 / 3.5 = 0.286. The closest answer provided is 0.284.

Question 9

A 0.200 mol sample of O₂ and 0.600 mol sample of H₂ are placed in a container at 1.20 atm and 25 ° C. What is the partial pressure of H₂ in this container?

Accepted Answer

The total moles of gas are 0.200 + 0.600 = 0.800 moles. The mole fraction of $H_2$ is $\frac{0.600}{0.800} = 0.75$. The partial pressure of $H_2$ is then $0.75 \times 1.20 \, \text{atm} = 0.90 \, \text{atm}$.

Question 10

Consider a mixture of gases consisting of 0.538 moles of He, 0.315 moles of Ne and 0.103 moles of Ar confined in a 7.00 L vessel at 25 ° C. What is the mole fraction of He gas in this mixture?

Accepted Answer

- First, find the total moles of gas in the mixture:

Total moles of gas = 0.538 mol He + 0.315 mol Ne + 0.103 mol Ar = 0.956 mol

- Next, calculate the mole fraction of He gas:

Mole fraction of He gas = moles of He gas / total moles of gas

Mole fraction of He gas = 0.538 mol / 0.956 mol = 0.563

Therefore, the mole fraction of He gas in the mixture is 0.563. The correct answer is D.

Question 11

What is the pressure in a 1.00 L container that contains 0.240 mol of N₂ and 0.360 mol of O₂ at a temperature of 27 ° C?

Accepted Answer

The answer of What is the pressure in a 1.00...

Question 12

A sample of gas is prepared by mixing 0.10 mol Ar, 0.10 mol N₂, and 0.30 mol He. If the entire vessel is at a pressure of 2.1 atm, what is the partial pressure of He?

Accepted Answer

The total number of moles of gas is 0.10 + 0.10 + 0.30 = 0.50 mol. 
The partial pressure of He can be calculated using the formula:

partial pressure of He = (moles of He / total moles of gas) x total pressure

Substituting the given values:

partial pressure of He = (0.30 / 0.50) x 2.1 atm 
partial pressure of He = 1.26 atm

Therefore, the closest option is B) 1.3 atm.

Question 13

Consider a mixture of gases consisting of 0.538 moles of He, 0.315 moles of Ne and 0.103 moles of Ar confined in a 7.00 L vessel at 25 ° C. What is the total pressure of this mixture of gases?

Accepted Answer

The total pressure of a mixture of gases can be calculated using the ideal gas law, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant (0.0821 L·atm/(K·mol)), and $T$ is the temperature in Kelvin. First, convert the temperature to Kelvin: $25 °C + 273.15 = 298.15 K$. Then, sum the moles of gases: $0.538 + 0.315 + 0.103 = 0.956$ moles. Using the ideal gas law: $P = \frac{nRT}{V} = \frac{0.956 \times 0.0821 \times 298.15}{7.00} ≈ 3.34$ atm.

Question 14

Consider a mixture of gases consisting of 0.538 moles of He, 0.315 moles of Ne and 0.103 moles of Ar confined in a 7.00 L vessel at 25 ° C. What is the partial pressure of the He gas?

Accepted Answer

The answer of Consider a mixture of gases consisting of...

Question 15

What is the partial pressure of Ar in a container that is at a total pressure of 1.23 atm and contains 2.3 g of Ar and 1.13 g of Ne?

Accepted Answer

The partial pressure of a gas is proportional to its mole fraction. Calculate moles of Ar (2.3 g / 39.95 g/mol) and Ne (1.13 g / 20.18 g/mol), find mole fraction of Ar, and multiply by total pressure (1.23 atm).

Question 16

A chemical engineer places a mixture of noble gases consisting of 10.0 grams of helium and 20.0 grams of Neon in a piston cylinder assembly. What is the partial pressure of helium in this mixture if the total pressure is 2.50 atm?

Accepted Answer

The answer of A chemical engineer places a mixture of...

Question 17

Exhibit 6-1 Consider a mixture of gases that includes 0.50 g of He, 5.0 g of Ne and 20 g of Ar contained in a 5.0 L vessel at 25 ° C to answer the following question(s). Molar Mass(He) = 4.0 g/mol, Molar Mass(Ne) = 20 g/mol, and Molar Mass(Ar) = 40 g/mol
-Refer to Exhibit 6-1. What is the total pressure of this mixture of gases?

Accepted Answer

The answer of Exhibit 6-1 Consider a mixture of gases...

Question 18

A 100 mL sample of oxygen gas is collected over water at 26 ° C and at a pressure of 756 torr. Given that the vapor pressure of water at 26 ° C is 25 torr, how many moles of dry oxygen are present?

Accepted Answer

The partial pressure of oxygen is 756 torr - 25 torr = 731 torr. The number of moles of oxygen is then:n = PV/RT = (731 torr * 100 mL) / (760 torr * 299 K * 0.0821 L atm / mol K) = 3.9 x 10^-3 mol

Question 19

What is the pressure in a 2.0 L container at 27 ° C that contains 2.0 g O₂ and 4.0 g N₂?

Accepted Answer

The answer of What is the pressure in a 2.0...

Question 20

What is the pressure in a 5.00 L container that contains 2.00 mol N₂ gas and 3.00 mol O₂ gas at 27 ° C?

Accepted Answer

The total moles of gas is 5.00 mol (2.00 mol N2 + 3.00 mol O2). Using the ideal gas law, PV = nRT, where R = 0.0821 L·atm/mol·K and T = 300 K (27 °C + 273), the pressure P = (5.00 mol * 0.0821 L·atm/mol·K * 300 K) / 5.00 L = 24.63 atm, approximately 24.6 atm.

Quiz 6: The Gaseous State