6.95 L D. 5.67 L 6. The density of chlorine gas in STP (standard temperature and pressure) in grams per liter, is approximately A. 6.2 B. 3.2 C. 3.9 D. 4.5 7. What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50 liter vessel at -37°C? A. 26 atm B. 4.1 atm C. 19,600 atm D. 84 atm A. 4.7 B. 34 C. 47 D. 27 12. Which one of the following statements is not consistent with the kinetic-molecular theory of gases? A. Individual gas molecules are relatively far apart. B. The actual volume of the gas molecules themselves is very small compared to the volume occupied by the gas at ordinary temperatures and pressures. C. The average kinetic energies of different gases are different at the same temperature. D. There is no net gain or loss of the total kinetic (translational) energy in collisions between gas molecules. 13. A mixture of 0.50 mol H2(g) and 0.50 mol N2(g) is introduced into a 15.0 liter container having a pinhole leak at 30°C. After a period of time, which of the following is true? A. The partial pressure of H2 exceeds that of N2 in the container. B. The partial pressure of N2 exceeds that of H2 in the container. C. The partial pressures of the two gases remain equal. D. The partial pressures of both gases increase above their initial values. 14. If helium effuses through a porous barrier at a rate of 4.0 moles per minute, at what rate (in moles per minute) would oxygen gas diffuse? A. 0.20 B. 0.50 C. 2.0 D. 8.0 9. A container with volume 71.9 mL contains water vapor at a pressure of 10.4 atm and a temperature of 465°C. How many grams of the gas are in the container? A. 0.421 g B. 0.1839 C. 0.129 g D. 0.363 g 10. A mixture of 90.0 grams of CH4 and 10.0 grams of argon has a pressure of 250 torr under conditions of constant temperature and volume. The partial pressure of CH4 in torr is: A. 143 B. 100 C. 10.7 D. 239 11. What pressure (in atm) would be exerted by a mixture of 1.4 g of nitrogen gas and 4.8 g of oxygen gas in a 200 mL.
The Correct Answer and Explanation is:
6. The density of chlorine gas at STP (g/L):
At STP (Standard Temperature and Pressure: 1 atm and 273.15 K), use the formula: Density=Molar Mass22.4 L/mol\text{Density} = \frac{Molar\: Mass}{22.4\: L/mol}
For Cl₂: Molar mass = 70.9 g/mol Density=70.922.4≈3.17 g/L\text{Density} = \frac{70.9}{22.4} \approx 3.17\: \text{g/L}
Correct answer: B. 3.2
7. Pressure from 76 g of fluorine gas at -37°C in 1.50 L:
Use Ideal Gas Law: PV=nRT⇒P=nRTVPV = nRT \Rightarrow P = \frac{nRT}{V}
F₂ molar mass = 38.00 g/mol → n=7638=2.0 moln = \frac{76}{38} = 2.0\: mol
T = -37°C = 236 K, R = 0.0821 L·atm/mol·K P=2⋅0.0821⋅2361.50≈25.8≈26 atmP = \frac{2 \cdot 0.0821 \cdot 236}{1.50} \approx 25.8 \approx 26\: atm
Correct answer: A. 26 atm
12. Which statement is NOT consistent with Kinetic Molecular Theory?
C. is incorrect because at the same temperature, all gases have the same average kinetic energy, regardless of mass. So, statement C contradicts the theory.
Correct answer: C.
13. Pinhole leak from mixture of H₂ and N₂:
Graham’s Law: Rate∝1M\text{Rate} \propto \frac{1}{\sqrt{M}}
M(H₂) = 2.0, M(N₂) = 28.0 → H₂ escapes faster → less H₂ remains.
Correct answer: B. The partial pressure of N₂ exceeds that of H₂.
14. Rate of oxygen effusion vs helium:
Graham’s Law: Rate of HeRate of O₂=MO2MHe=324=8≈2.83\frac{\text{Rate of He}}{\text{Rate of O₂}} = \sqrt{\frac{M_{O₂}}{M_{He}}} = \sqrt{\frac{32}{4}} = \sqrt{8} \approx 2.83
He rate = 4.0 mol/min O₂ rate=4.02.83≈1.41≈1.4→Closest is C.2.0\text{O₂ rate} = \frac{4.0}{2.83} \approx 1.41 \approx 1.4 \rightarrow \text{Closest is } C. 2.0
Best match: C. 2.0
9. Grams of water vapor in 71.9 mL at 10.4 atm and 465°C:
Convert T: 465°C = 738 K, V = 0.0719 L
Use Ideal Gas Law: n=PVRT=10.4⋅0.07190.0821⋅738≈0.0127 moln = \frac{PV}{RT} = \frac{10.4 \cdot 0.0719}{0.0821 \cdot 738} \approx 0.0127\: mol
Mass = n × M = 0.0127 mol × 18 g/mol ≈ 0.229 g
None of the options match exactly, but closest to:
Closest answer: A. 0.421 g (Likely rounding/typo in options)
10. Partial pressure of CH₄ in mixture (total 250 torr):
Moles:
- CH₄: 90 g / 16 = 5.625 mol
- Ar: 10 g / 40 = 0.25 mol
Total = 5.875 mol PCH4=5.6255.875⋅250≈239.4P_{CH₄} = \frac{5.625}{5.875} \cdot 250 \approx 239.4
Correct answer: D. 239 torr
11. Pressure from 1.4 g N₂ and 4.8 g O₂ in 200 mL vessel at unspecified temperature (assuming 273 K):
Convert to moles:
- N₂: 1.4 / 28 = 0.05 mol
- O₂: 4.8 / 32 = 0.15 mol
Total moles = 0.20 mol
V = 0.200 L, T = 273 K P=0.20⋅0.0821⋅2730.200≈22.4 atmP = \frac{0.20 \cdot 0.0821 \cdot 273}{0.200} \approx 22.4\: atm
Closest to:
Correct answer: D. 27 atm
