Pre AP Chemistry -2.1C Partial Pressure and Gas Mixtures- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Chemistry -2.1C Partial Pressure and Gas Mixtures- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Chemistry -2.1C Partial Pressure and Gas Mixtures- MCQ Exam Style Questions – Pre AP Chemistry – per latest Pre AP Chemistry Syllabus.
Two separate flasks are connected. Flask A (1.0 L) contains \(N_2\) at 0.60 atm. Flask B (2.0 L) contains \(O_2\) at 0.30 atm. When the stopcock is opened and the gases mix at constant temperature, what is the total pressure?
(A) 0.20 atm
(B) 0.40 atm
(C) 0.45 atm
(D) 0.90 atm
▶️ Answer / Explanation
Total volume after mixing:
\(V = 1.0 + 2.0 = 3.0\) L
New pressure of \(N_2\):
\(P = \frac{0.60 \times 1.0}{3.0} = 0.20\) atm
New pressure of \(O_2\):
\(P = \frac{0.30 \times 2.0}{3.0} = 0.20\) atm
Total pressure:
\(P_{total} = 0.20 + 0.20 = 0.40\) atm
Answer: B
A gas mixture contains 40% \(H_2\), 35% \(N_2\), and 25% \(CO_2\) by moles. If the total pressure is 2.00 atm, which gas has the highest partial pressure and what is its value?
(A) \(H_2\), 0.80 atm
(B) \(N_2\), 0.70 atm
(C) \(H_2\), 0.40 atm
(D) \(CO_2\), 0.50 atm
▶️ Answer / Explanation
Partial pressure = mole fraction × total pressure.
\(P_{H_2} = 0.40 \times 2.00 = 0.80\) atm
\(P_{N_2} = 0.35 \times 2.00 = 0.70\) atm
\(P_{CO_2} = 0.25 \times 2.00 = 0.50\) atm
The highest value is for \(H_2\).
Answer: A
A student collects hydrogen gas over water at 30°C. The total pressure is 745 mmHg and the vapor pressure of water is 31.8 mmHg. The volume of gas collected is 250 mL. Approximately how many moles of dry \(H_2\) were produced? (R = 62.4 L·mmHg·mol⁻¹·K⁻¹)
(A) \(9.1 \times 10^{-3}\) mol
(B) \(9.5 \times 10^{-3}\) mol
(C) \(1.0 \times 10^{-2}\) mol
(D) \(1.2 \times 10^{-2}\) mol
▶️ Answer / Explanation
Dry hydrogen pressure:
\(P_{H_2} = 745 – 31.8 = 713.2\) mmHg
Using ideal gas equation:
\(n = \frac{PV}{RT}\)
\(n = \frac{713.2 \times 0.250}{62.4 \times 303}\)
\(n \approx 9.4 \times 10^{-3}\) mol
Closest option:
Answer: B
A sealed rigid container holds a mixture of \(CH_4\) and \(C_2H_6\). The total pressure is 1.50 atm, and the mole fraction of \(CH_4\) is 0.60. If the temperature is increased so the total pressure becomes 3.00 atm, what is the new partial pressure of \(C_2H_6\)?
(A) 0.60 atm
(B) 0.90 atm
(C) 1.20 atm
(D) 1.80 atm
▶️ Answer / Explanation
Mole fraction of \(C_2H_6\):
\(1 – 0.60 = 0.40\)
In a rigid container the mole fractions remain constant when temperature changes.
New partial pressure:
\(P = 0.40 \times 3.00 = 1.20\) atm
Answer: C
A container holds a mixture of nitrogen gas and oxygen gas. The total pressure is 1.20 atm. If the partial pressure of nitrogen is 0.80 atm, what is the mole fraction of oxygen in the mixture?
A. 0.33
B. 0.40
C. 0.67
D. 0.80
▶️ Answer / Explanation
Using Dalton’s Law:
\(P_{O_2} = P_{total} – P_{N_2}\)
\(P_{O_2} = 1.20 – 0.80 = 0.40\) atm
Mole fraction:
\(\chi_{O_2} = \dfrac{P_{O_2}}{P_{total}} = \dfrac{0.40}{1.20} = 0.33\)
Answer: A
A gas mixture contains 2.0 mol He, 3.0 mol Ne, and 5.0 mol Ar. What is the partial pressure of neon if the total pressure is 500 mmHg?
A. 100 mmHg
B. 150 mmHg
C. 167 mmHg
D. 250 mmHg
▶️ Answer / Explanation
Total moles:
\(2.0 + 3.0 + 5.0 = 10.0\) mol
Mole fraction of Ne:
\(\chi_{Ne} = \dfrac{3}{10} = 0.30\)
Partial pressure:
\(P_{Ne} = 0.30 \times 500 = 150\) mmHg
Answer: B
A gas is collected over water at 25°C. The total pressure is 760 mmHg. If the vapor pressure of water at 25°C is 23.8 mmHg, what is the partial pressure of the dry gas?
A. 23.8 mmHg
B. 736.2 mmHg
C. 760.0 mmHg
D. 783.8 mmHg
▶️ Answer / Explanation
\(P_{dry} = P_{total} – P_{H_2O}\)
\(P_{dry} = 760 – 23.8 = 736.2\) mmHg
Answer: B
Which statement best explains why Dalton’s Law of Partial Pressures applies to ideal gas mixtures?
(A) Ideal gas molecules have negligible volume and no intermolecular forces.
(B) All gas molecules have identical masses.
(C) Heavier molecules settle at the bottom of the container.
(D) Gas molecules collide more frequently in mixtures.
▶️ Answer / Explanation
Dalton’s Law works because ideal gases:
• occupy negligible volume
• have no intermolecular attractions
Therefore each gas behaves independently and contributes its own pressure.
Answer: A
A 4.0 L container at 300 K contains 0.50 mol \(N_2\) and 0.25 mol \(O_2\). After a reaction, the partial pressure of \(O_2\) becomes 0.40 atm. What is the new total pressure in the container?
(A) 1.23 atm
(B) 1.54 atm
(C) 1.94 atm
(D) 2.47 atm
▶️ Answer / Explanation
Pressure of \(N_2\) using ideal gas equation:
\(P = \frac{nRT}{V}\)
\(P_{N_2} = \frac{0.50 \times 0.0821 \times 300}{4.0}\)
\(P_{N_2} \approx 1.54\) atm
Total pressure:
\(P_{total} = 1.54 + 0.40 = 1.94\) atm
Answer: C
Flask X (3.0 L) contains \(CO_2\) at 2.0 atm and Flask Y (1.0 L) contains \(He\) at 4.0 atm. The gases are mixed at constant temperature. What is the mole fraction of \(CO_2\) in the final mixture?
(A) 0.25
(B) 0.50
(C) 0.60
(D) 0.75
▶️ Answer / Explanation
Moles are proportional to \(PV\) when temperature is constant.
\(CO_2: 2.0 \times 3.0 = 6\)
\(He: 4.0 \times 1.0 = 4\)
Total = 10
Mole fraction of \(CO_2\):
\(\frac{6}{10} = 0.60\)
Answer: C
A mixture of \(H_2\) and \(Cl_2\) gases is placed in a rigid container. The total pressure is 1.80 atm and the gases are present in equal mole fractions. They react completely according to:
\(H_2 + Cl_2 \rightarrow 2HCl\)
What is the final total pressure at constant temperature?
(A) 0.90 atm
(B) 1.80 atm
(C) 2.70 atm
(D) 3.60 atm
▶️ Answer / Explanation
Assume each gas has \(n\) moles.
Initial total moles = \(2n\)
Reaction:
\(H_2 + Cl_2 \rightarrow 2HCl\)
\(n\) mol \(H_2\) + \(n\) mol \(Cl_2\) → \(2n\) mol \(HCl\)
Total moles remain \(2n\).
Since pressure is proportional to number of moles at constant temperature and volume, the pressure remains unchanged.
Final pressure = 1.80 atm
Answer: B
A rigid 10.0 L vessel contains gases with the following partial pressures:
\(P_{Ar} = 0.30\) atm
\(P_{N_2} = 0.50\) atm
\(P_{CO_2} = 0.20\) atm
If the temperature increases from 27°C to 327°C, what is the new partial pressure of \(N_2\)?
(A) 0.50 atm
(B) 0.75 atm
(C) 1.00 atm
(D) 1.50 atm
▶️ Answer / Explanation
Convert temperatures to Kelvin:
\(T_1 = 300K\)
\(T_2 = 600K\)
Using Gay-Lussac’s law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
\(P_2 = P_1 \times \frac{T_2}{T_1}\)
\(P_2 = 0.50 \times \frac{600}{300}\)
\(P_2 = 1.00\) atm
Answer: C
Oxygen gas is collected over water at 20°C. The volume of gas collected is 310 mL at a total pressure of 755 mmHg. The vapor pressure of water at 20°C is 17.5 mmHg.
What mass of \(O_2\) was collected? (R = 62.4 L·mmHg·mol⁻¹·K⁻¹)
(A) 0.118 g
(B) 0.392 g
(C) 0.402 g
(D) 0.422 g
▶️ Answer / Explanation
Dry oxygen pressure:
\(P_{O_2} = 755 – 17.5 = 737.5\) mmHg
Using ideal gas equation:
\(n = \frac{PV}{RT}\)
\(n = \frac{737.5 \times 0.310}{62.4 \times 293}\)
\(n ≈ 0.0125\) mol
Mass of oxygen:
\(m = n \times M\)
\(m = 0.0125 \times 32\)
\(m ≈ 0.40 g\)
Closest answer:
Answer: B
A 2.0 L flask contains \(N_2\) at 3.0 atm and a 6.0 L flask contains \(Ar\) at 1.0 atm. The gases are mixed at constant temperature.
What is the final total pressure?
(A) 0.75 atm
(B) 1.35 atm
(C) 1.50 atm
(D) 4.00 atm
▶️ Answer / Explanation
Total volume:
\(2 + 6 = 8\) L
New pressure of \(N_2\):
\(\frac{3.0 \times 2}{8} = 0.75\) atm
New pressure of \(Ar\):
\(\frac{1.0 \times 6}{8} = 0.75\) atm
Total pressure:
\(0.75 + 0.75 = 1.50\) atm
Answer: C
A scuba tank contains a gas mixture with 36% \(O_2\) and 64% \(N_2\) by moles at a pressure of 200 atm. A diver breathes this mixture at a depth where the ambient pressure is 4.0 atm.
What is the partial pressure of \(O_2\) the diver breathes?
(A) 0.84 atm
(B) 1.44 atm
(C) 2.56 atm
(D) 72.0 atm
▶️ Answer / Explanation
The regulator supplies gas at the ambient pressure (4 atm).
Mole fraction of oxygen:
\(X_{O_2} = 0.36\)
Partial pressure:
\(P_{O_2} = X_{O_2} \times P_{total}\)
\(P_{O_2} = 0.36 \times 4\)
\(P_{O_2} = 1.44\) atm
Answer: B