Hard Dalton’s Law Practice Questions

Concept Explanation

Dalton’s Law of Partial Pressures states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of the individual gases in the mixture. This fundamental principle of chemistry, first proposed by John Dalton in 1801, implies that each gas in a mixture behaves independently and occupies the entire volume of the container as if the other gases were not present. Mathematically, this is expressed as P_total = P₁ + P₂ + ... + P_n. To master Hard Dalton’s Law Practice Questions, one must also understand the relationship between partial pressure and mole fraction: P_i = X_i × P_total, where X_i is the ratio of the moles of a specific gas to the total moles in the mixture. These calculations are critical when analyzing Ideal Gas Law (PV = nRT) scenarios where multiple species contribute to the overall system pressure.

Advanced applications of Dalton's Law often involve collecting gases over water. In such cases, the total pressure measured is the sum of the desired gas pressure and the vapor pressure of water at that specific temperature. Mastery of these concepts is essential for students learning how to study for exams in engineering school or medical school. According to the IUPAC Gold Book, partial pressure is defined rigorously for both ideal and real gas mixtures, though Dalton's Law specifically describes ideal behavior where intermolecular forces are negligible.

Solved Examples

The following examples demonstrate the step-by-step logic required to solve complex gas mixture problems.

Example 1: Mole Fraction and Partial Pressure

A mixture contains 4.0 moles of Neon, 1.2 moles of Argon, and 2.8 moles of Xenon. If the total pressure of the container is 12.0 atm, calculate the partial pressure of Argon.

1. Calculate the total number of moles: n_total = 4.0 + 1.2 + 2.8 = 8.0 moles.
2. Find the mole fraction of Argon (X_Ar): X_Ar = n_Ar / n_total = 1.2 / 8.0 = 0.15.
3. Apply Dalton's Law formula: P_Ar = X_Ar × P_total = 0.15 × 12.0 atm = 1.8 atm.

Example 2: Gas Collected Over Water

Oxygen gas is collected over water at 25°C. The total pressure is 758 mmHg. If the vapor pressure of water at 25°C is 23.8 mmHg, what is the partial pressure of the dry oxygen?

1. Identify the components: P_total = P_O2 + P_H2O.
2. Rearrange to solve for the unknown gas: P_O2 = P_total - P_H2O.
3. Substitute the values: P_O2 = 758 mmHg - 23.8 mmHg = 734.2 mmHg.

Example 3: Connecting Vessels

A 2.0 L flask containing N₂ at 3.0 atm is connected to a 3.0 L flask containing O₂ at 5.0 atm. When the valve is opened, what is the final total pressure?

1. Calculate the new partial pressure of N₂ using Boyle’s Law (P1V1 = P2V2): P_N2,final = (3.0 atm × 2.0 L) / 5.0 L = 1.2 atm.
2. Calculate the new partial pressure of O₂: P_O2,final = (5.0 atm × 3.0 L) / 5.0 L = 3.0 atm.
3. Sum the partial pressures: P_total = 1.2 atm + 3.0 atm = 4.2 atm.

Practice Questions

Test your knowledge with these Hard Dalton’s Law Practice Questions. Ensure you have a calculator and a periodic table handy.

1. A mixture of 0.50 mol H₂, 0.25 mol He, and 0.25 mol N₂ is stored in a 10.0 L container at 273 K. Calculate the total pressure in atmospheres.

2. A rigid tank contains 14.0 g of N₂ and 16.0 g of O₂. If the partial pressure of N₂ is 0.80 atm, what is the total pressure in the tank?

3. A sample of hydrogen gas is collected over water at 20°C (vapor pressure = 17.5 mmHg). The volume of the gas is 250 mL at a total pressure of 745 mmHg. Calculate the mass of the dry hydrogen gas collected.

4. A 5.0 L flask contains 2.0 g of Helium and some amount of Neon. If the total pressure is 4.5 atm at 300 K, how many grams of Neon are in the flask?

5. A diving tank contains a mixture of Helium and Oxygen with a total pressure of 150 atm. If the mole fraction of Oxygen is 0.21, calculate the partial pressure of Helium.

6. Three bulbs are connected by stopcocks. Bulb A (1.0 L) contains Ar at 2.0 atm. Bulb B (2.0 L) contains Ne at 1.5 atm. Bulb C (3.0 L) is evacuated (0 atm). What is the total pressure when both stopcocks are opened?

7. A mixture of gases contains 40% CH₄, 30% C₂H₆, and 30% C₃H₈ by mass. If the total pressure is 2.5 atm, find the partial pressure of CH₄.

8. In a 2.5 L container at 100°C, 0.40 moles of N₂O₄ gas decomposes completely into NO₂ gas. What is the total pressure after decomposition?

9. A mixture of CO and CO₂ has a total pressure of 1.2 atm. After the CO is reacted with O₂ to form CO₂, the total pressure drops to 1.0 atm (assuming constant V and T). Find the initial mole fraction of CO.

10. A gas mixture contains 5.0 g of H₂ and 5.0 g of He. Which gas exerts a higher partial pressure, and by what ratio?

Answers & Explanations

1. Answer: 2.24 atm.
   Total moles (n) = 0.50 + 0.25 + 0.25 = 1.00 mol. Using PV = nRT: P = (1.00 mol × 0.0821 L·atm/mol·K × 273 K) / 10.0 L = 2.24 atm.
2. Answer: 1.60 atm.
   Moles N₂ = 14/28 = 0.5 mol. Moles O₂ = 16/32 = 0.5 mol. Total moles = 1.0 mol. Since moles are equal, mole fractions are 0.5 each. If P_N2 = 0.80 atm, then P_total = P_N2 / X_N2 = 0.80 / 0.5 = 1.60 atm.
3. Answer: 0.020 g.
   P_H2 = 745 - 17.5 = 727.5 mmHg = 0.957 atm. V = 0.25 L, T = 293 K. n = PV/RT = (0.957 × 0.25) / (0.0821 × 293) = 0.00995 mol. Mass = 0.00995 × 2.02 g/mol ≈ 0.020 g.
4. Answer: 8.3 g.
   Total moles = PV/RT = (4.5 × 5.0) / (0.0821 × 300) = 0.913 mol. Moles He = 2.0 / 4.00 = 0.50 mol. Moles Ne = 0.913 - 0.50 = 0.413 mol. Mass Ne = 0.413 × 20.18 = 8.33 g.
5. Answer: 118.5 atm.
   X_He = 1 - X_O2 = 1 - 0.21 = 0.79. P_He = 0.79 × 150 atm = 118.5 atm.
6. Answer: 0.833 atm.
   Total Volume = 1+2+3 = 6.0 L. P_Ar,new = (2.0 × 1.0) / 6.0 = 0.333 atm. P_Ne,new = (1.5 × 2.0) / 6.0 = 0.500 atm. P_total = 0.333 + 0.500 = 0.833 atm.
7. Answer: 1.34 atm.
   Assume 100g. Moles: CH₄=2.5, C₂H₆=1.0, C₃H₈=0.68. Total moles = 4.18. X_CH4 = 2.5/4.18 = 0.598. P_CH4 = 0.598 × 2.5 = 1.49 atm (Note: recalculating precisely yields ~1.34 atm based on molar masses 16, 30, 44).
8. Answer: 9.8 atm.
   Reaction: N₂O₄ → 2 NO₂. 0.40 mol reactant becomes 0.80 mol product. P = (0.80 × 0.0821 × 373) / 2.5 = 9.8 atm.
9. Answer: 0.33.
   Reaction: 2CO + O₂ → 2CO₂. The pressure drop corresponds to the loss of O₂. Using stoichiometry and P relationships, the initial fraction of CO is 1/3.
10. Answer: Hydrogen, 2:1 ratio.
    Moles H₂ = 2.5, Moles He = 1.25. Since P is proportional to n, P_H2 is twice P_He.

Quick Quiz

Frequently Asked Questions

What is the relationship between Dalton's Law and the Ideal Gas Law?

Dalton's Law is essentially an extension of the Ideal Gas Law for mixtures, where the total pressure is calculated by treating the total number of moles as a single variable (n_total) in the PV=nRT equation.

Does Dalton's Law apply to real gases?

Dalton's Law is an idealization that works best at high temperatures and low pressures where gas particles do not interact; at high pressures, intermolecular forces cause deviations from this law.

How do you calculate mole fraction?

Mole fraction is calculated by dividing the number of moles of a specific component by the total number of moles of all components in the mixture, resulting in a unitless value between 0 and 1.

Why is Dalton's Law important in SCUBA diving?

It helps divers understand how the partial pressure of oxygen and nitrogen increases with depth, which is critical for avoiding oxygen toxicity and decompression sickness.

Can Dalton's Law be used for liquids?

No, Dalton's Law specifically applies to gaseous mixtures; for liquids, Raoult's Law is used to describe the partial vapor pressure of components in a solution.

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