Chapter 17: Gases
Questions (with solutions!)

NOTE: The solutions are wrote by ChatGPT.
I am aware of the solutions may be bad/confusing (I tried to edit the ones that may be confusing). If you are still confused on any question, feel free to ask someone that's good at the topic (or ask ChatGPT for more clarification lol) or you can just skip the question for now, and come back later.
This also means that the solution MAY be wrong (or I wrote the questions wrong for ChatGPT).

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1. 0.1 mol of metal M reacts completely with 1.68 dm³ of oxygen gas at STP. What is the molecular formula of the oxide formed?
A. M₃O₄    B. M₂O₃    C. M₂O    D. MO

Solution:
At STP, 1 mol of gas = 22.4 dm³
\[ \text{Moles of } O_2 = \frac{1.68}{22.4} = 0.075 \text{ mol} \]
Each O₂ molecule gives 2 O atoms:
\[ 0.075 \text{ mol } O_2 \times 2 = 0.15 \text{ mol O atoms} \]
Given: 0.1 mol of M
Ratio M : O = 0.1 : 0.15 → simplify:
\[ \frac{0.1}{0.05} : \frac{0.15}{0.05} = 2 : 3 \]
Empirical formula is M₂O₃
Answer: B. M₂O₃

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2. At room temperature, 1 dm³ of ammonia at 1 atm and 1 dm³ of hydrogen chloride at 2 atm are introduced into a 2 dm³ vessel. What is the total pressure of the gases in the vessel?
A. 0.5 atm    B. 1 atm    C. 1.5 atm    D. 3 atm

Solution:
Use Boyle's Law: \( P_1V_1 = P_2V_2 \)
For NH₃:
\[ P_{NH3} = \frac{1 \times 1}{2} = 0.5 \text{ atm} \]
For HCl:
\[ P_{HCl}= \frac{2 \times 1}{2} = 1 \text{ atm} \]
Total pressure = 0.5 + 1 = 1.5 atm
Answer: C. 1.5 atm

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3. It takes 8 seconds to diffuse 80 cm³ of oxygen through a small hole. Under the same conditions, 40 cm³ of gas K takes 10 seconds to diffuse. What is the molecular weight of gas K?
A. 20.48    B. 80    C. 200    D. 250

Solution:
Use Graham's Law:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
Rate of O₂: \( \frac{80}{8} = 10 \text{ cm³/s} \)
Rate of K: \( \frac{40}{10} = 4 \text{ cm³/s} \)
\[ \frac{10}{4} = \sqrt{\frac{M_K}{32}} \] \[ 2.5 = \sqrt{\frac{M_K}{32}} \]
Square both sides:
\[ 6.25 = \frac{M_K}{32} \] \[ M_K = 6.25 \times 32 = 200 \]
Answer: C. 200

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4. Which of the following represents 1 mole of a substance?
I. 8 g of carbon
II. 16 g of sulfur
III. 32 g of oxygen
IV. 22.4 dm³ of oxygen at STP
V. \( 6.023 \times 10^{23} \) helium molecules
A. III, V  B. I, II, IV  C. II, III, V  D. III, IV, V

Answer:
- 8 g of carbon (C = 12) → \( \frac{8}{12} = 0.67 \) mol → not 1 mol
- 16 g of sulfur (S = 32) → \( \frac{16}{32} = 0.5 \) mol → not 1 mol
- 32 g of O₂ → 1 mol ✅
- 22.4 dm³ of oxygen gas at STP = 1 mol ✅
- \( 6.023 \times 10^{23} \) helium atoms = 1 mol ✅
Answer: D. III, IV, V

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5. The relative molecular masses of gases A and B are 2 and 50 respectively. Under the same conditions, gas A diffuses at a rate of ___ times that of gas B.
A. \(\frac{1}{25}\)  B. 5  C. 25  D. 100

Solution:
Use Graham's Law:
\[ \frac{r_A}{r_B} = \sqrt{ \frac{M_B}{M_A} } \] \[ = \sqrt{ \frac{50}{2} } = \sqrt{25} = 5 \]
Answer: B. 5

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6. Which of the following statements about the kinetic molecular theory of gases is correct?
I. The number of gas molecules increases when the temperature increases
II. In a spherical container, gas molecules move curvilinearly
III. At room temperature, all gases can be liquefied
IV. At constant temperature, gas pressure is inversely proportional to its volume
V. As the absolute temperature increases, the average kinetic energy of gas molecules increases
A. I, IV  B. IV, V  C. II, III, V  D. II, III, IV, V

Answer:
I. ❌ Number of molecules doesn't increase with temp (only kinetic energy does)
II. ❌ Molecules move in straight lines (not curves)
III. ❌ Not all gases can be liquefied at room temperature
IV. ✅ Boyle's Law
V. ✅ Kinetic energy ∝ absolute temperature
Answer: B. IV, V

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7. A container is filled with a certain mass of gas. If the volume is decreased at constant temperature, which of the following will NOT be affected?
I. The pressure in the container
II. The speed of the molecules
III. The average kinetic energy of the molecules
IV. The frequency of collisions with the container wall
A. III  B. I, IV  C. II, III  D. II, IV

Answer:
I. ✅ Affected (pressure increases)
II. ❌ Speed depends on temperature (not affected if temp constant)
III. ✅ Kinetic energy depends only on temp → not affected
IV. ✅ Collisions increase when volume decreases
Answer: A. III

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8. Under the same temperature and pressure, the ratio of the volume of O₂ to that of CO₂ (same mass) would be ___
A. 8:11  B. 5:6  C. 11:8  D. 11:4

Solution:
Let mass = m
Molar mass of O₂ = 32, CO₂ = 44
Moles = mass / molar mass → volume ∝ moles
So: \[ \text{Volume ratio} = \frac{\frac{m}{32}}{\frac{m}{44}} = \frac{44}{32} = \frac{11}{8} \]
Answer: C. 11:8

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9. Under the same temperature and pressure, two gases have the same volume. They surely ___.
I React with each other in the same volume
II Contain the same number of molecules
III Contain the same number of atoms
IV Have the same number of moles
A. I, II  B. I, III  C. II, IV  D. III, IV

Solution:
At same temperature and pressure, equal volumes of gases contain equal number of molecules (Avogadro's Law).
II. ✅ True (Avogadro's Law)
IV. ✅ True (since mole = number of molecules / Avogadro constant)
I and III ❌ Not always true
Answer: C. II, IV

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10. Under the same temperature and pressure, 100 mL of P₂ combines with 50 mL of Q₂ to form 100 mL of R. What is the molecular formula of R?
A. PQ  B. PQ₂  C. P₂Q  D. P₂Q₂

Solution:
Volume ratio:
P₂ : Q₂ : R = 100 : 50 : 100 = 2 : 1 : 2
Thus 2 vol P₂ + 1 vol Q₂ → 2 vol R
This gives molecular formula: P₂Q
Answer: C. P₂Q

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11. When 5 g of impure zinc powder reacts with excess hydrochloric acid, 1500 cm³ of hydrogen gas is collected at STP. What is the purity of the zinc powder?
A. 43.5%  B. 66.9%  C. 74.7%  D. 87.1%

Solution:
STP: 1 mol gas = 22400 cm³ \[ n(H_2) = \frac{1500}{22400} = 0.06696 \text{ mol} \] From Zn + 2HCl → ZnCl₂ + H₂, 1 mol Zn → 1 mol H₂ \[ \text{Mass of Zn reacted} = 0.06696 \times 65.38 = 4.37 \text{ g} \] \[ \text{Purity} = \frac{4.37}{5.00} \times 100\% = 87.4\% \] Closest option: 87.1% (This question may be poorly designed OR just outdated. The more accurate answer should be 87.56%.)
Answer: D. 87.1%

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12. Which of the following has the most number of atoms at room temperature and pressure?
A. 7 g of hydrogen
B. 3 mol of ammonia gas
C. 2 mol of nitrogen gas
D. 88 dm³ of carbon dioxide

Solution:
A. 7 g H₂ = 3.5 mol H₂ → 3.5 × 2 = 7 mol atoms
B. 3 mol NH₃ → 3 × (1 N + 3 H) = 3 × 4 = 12 mol atoms
C. 2 mol N₂ → 4 mol atoms
D. 88 dm³ CO₂ at RTP = 88 / 24 = 3.67 mol → 3.67 × 3 = 11 mol atoms
Most atoms = option B (12 mol atoms)
Answer: B. 3 mol of ammonia gas

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13. Which of the following substances contains the most number of atoms?
A. 4 g of hydrogen gas
B. 18 mL water at 4 °C
C. 22.4 L helium gas at STP
D. 11.2 L oxygen gas at STP

Solution:
A. H₂: 4 g → 2 mol → 4 mol atoms
B. Water: 18 mL = 18 g = 1 mol → 3 mol atoms
C. 22.4 L He = 1 mol atoms
D. 11.2 L O₂ = 0.5 mol O₂ = 1 mol atoms
Most atoms = A (4 mol)
Answer: A. 4 g of hydrogen gas

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14. At 27 °C and \( 2.5 \times 10^4 \) Pa, the volume of 12 g of a gas is 74.82 L. What is this gas?
A. C₂H₂  B. CH₄  C. CO₂  D. CO

Solution:
Use ideal gas law: \[ PV = nRT \Rightarrow n = \frac{PV}{RT} \] T = 27 + 273 = 300 K \[ n = \frac{2.5 \times 10^4 \times 74.82}{8.314 \times 300} ≈ 0.75 \text{ mol} \] Molar mass = \( \frac{12}{0.75} = 16 gmol^{-1} \) → CH₄
Answer: B. CH₄

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15. Under what conditions does a real gas behave most like an ideal gas?
I Near the boiling point
II At high temperature
III At low temperature
IV At high pressure
V At low pressure
A. I, II  B. II, V  C. III, IV  D. I, IV, V

Solution:
Real gases behave most ideally at high temperature and low pressure (particles far apart).
I. ❌ Near boiling point → gas tends to liquefy
II. ✅ High temperature → behaves more ideally
V. ✅ Low pressure → less interaction
Answer: B. II, V

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16. A gas sample contains 10 g of neon and 8 g of oxygen. The total pressure of this gas mixture at 25 °C is 100 kPa. What is the partial pressure of oxygen?
A. 33 kPa  B. 50 kPa  C. 55 kPa  D. 67 kPa

Solution:
Moles of Ne = \( \frac{10}{20.18} ≈ 0.495 \) mol
Moles of O₂ = \( \frac{8}{32} = 0.25 \) mol
Total mol = 0.495 + 0.25 = 0.745 mol
\[ P_{O_2} = \frac{0.25}{0.745} \times 100 = 33.56 \text{ kPa} \]
Answer: A. 33 kPa

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17. How much pressure is exerted by a mixture of 16 g of oxygen and 7 g of nitrogen in a 5 L container at 20 °C?
A. \( 7.31 \times 10^5 \) Pa
B. \( 3.65 \times 10^5 \) Pa
C. \( 4.99 \times 10^4 \) Pa
D. \( 1.87 \times 10^4 \) Pa

Solution:
O₂: 16 g → 0.5 mol  N₂: 7 g → 0.25 mol  Total mol = 0.75 mol
T = 20 + 273 = 293 K
Use ideal gas law:
\[ P = \frac{nRT}{V} = \frac{0.75 \times 8.314 \times 293}{5} = 365.2 \times 10^3 \text{ Pa} \] \[ P = \frac{0.75 \times 8.314 \times 293}{5} \] \[ P = 365.2 \times 10^3 \text{ Pa} \]
Answer: B. \( 3.65 \times 10^5 \) Pa

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18. Which of the following are characteristics of an ideal gas?
I No intermolecular forces
II No volume of molecules
III No mass of molecules
A. I, II  B. I, III  C. II, III  D. I, II, III

Solution:
I. ✅ Ideal gas has no intermolecular forces
II. ✅ Assumes zero volume of particles
III. ❌ Molecules have mass (needed for kinetic energy)
Answer: A. I, II

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19. Which of the following contains the greatest number of atoms?
A. 10 g of neon
B. 0.4 mol of oxygen
C. 5.4 mL of water at 4 °C
D. 5.6 L of carbon dioxide at STP

Solution:
A. Ne: 10 g / 20 = 0.5 mol atoms
B. 0.4 mol O₂ = 0.8 mol atoms
C. Water: 5.4 mL ≈ 0.3 mol → 0.9 mol atoms
D. 5.6 L CO₂ = 0.25 mol → 0.75 mol atoms
→ C has most atoms
Answer: C. 5.4 mL of water at 4 °C

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20. A mixture of 20 cm³ oxygen and 30 cm³ hydrogen explodes and burns, forming water. What is the volume of the remaining gas?
A. 5 cm³  B. 15 cm³  C. 20 cm³  D. 30 cm³

Solution:
2H₂ + O₂ → 2H₂O (volume ratio 2:1)
30 cm³ H₂ reacts with 15 cm³ O₂ → only 20 cm³ O₂ available
O₂ is limiting → reacts with 40 cm³ H₂ → excess = 30 - 40 = -10 → can't happen
H₂ is limiting → 30 cm³ H₂ needs 15 cm³ O₂
Remaining O₂ = 20 - 15 = 5 cm³
Answer: A. 5 cm³

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21. Arrange the following gases in order of decreasing density at STP:
I. O₂
II. SO₂
III. CO₂
IV. NO₂
A. I → III → II → IV
B. II → IV → III → I
C. III → II → IV → I
D. IV → I → II → III

Solution:
Density ∝ molar mass
M(O₂) = 32  CO₂ = 44  NO₂ = 46  SO₂ = 64
Descending: SO₂ > NO₂ > CO₂ > O₂
Answer: B. II → IV → III → I

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22. A tyre took 20 minutes to deflate when filled with air (mean relative molecular mass = 28.8). How long will it take to deflate if filled with hydrogen at the same temperature and pressure?
A. \( \frac{20 \times 28.8}{2} \) min
B. \( 20 \times \sqrt{ \frac{28.8}{2} } \) min
C. \( 20 \times \sqrt{ \frac{2}{28.8} } \) min
D. \( \frac{20 \times 2}{28.8} \) min

Solution:
Rate ∝ \( \frac{1}{\sqrt{M}} \) → time ∝ \( \sqrt{M} \)
Let \( t_1 = 20 \), \( M_1 = 28.8 \), \( M_2 = 2 \)
\[ t_2 = 20 \times \sqrt{ \frac{28.8}{2} } \]
Answer: B. \( 20 \times \sqrt{ \frac{28.8}{2} } \) min

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23. Two vessels of equal volume contain NO₂ gas and a mixture of N₂ and O₂ gas with equal number of moles. Under the same temperature and pressure, which of the following must be equal for the two vessels?
A. Mass
B. Number of moles
C. Total number of atoms
D. Total number of protons

Solution:
Same number of moles → same number of molecules
Mass will differ due to different molar masses
Atoms: NO₂ has 3 atoms/molecule; N₂ + O₂ mixture averages less
Protons depend on atomic identity, not number of molecules
Only number of moles is guaranteed to be equal
Answer: B. Number of moles

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24. A nitrogen-hydrogen mixture, initially in the mole ratio of 1:3, reaches equilibrium with ammonia when 50% of the material has reacted. If the total final pressure is P, what is the partial pressure of ammonia?
A. \( \frac{P}{3} \)  B. \( \frac{P}{4} \)  C. \( \frac{P}{6} \)  D. \( \frac{P}{8} \)

Solution:
N₂ + 3H₂ ⇌ 2NH₃
Initial: 1 mol N₂, 3 mol H₂
50% reacted: 0.5 mol N₂ used → 1.5 mol H₂ used → 1 mol NH₃ formed
Total final mol = 0.5 N₂ + 1.5 H₂ + 1 NH₃ = 3 mol
\[ \text{Partial pressure of NH₃} = \frac{1}{3}P \]
Answer: A. \( \frac{P}{3} \)

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25. At 273 °C and \(1.01 \times 10^5\ \text{Pa}\), 1.4 g of nitrogen and 4.0 g of argon are mixed. What is the volume of this gas mixture?
A. 3.37 L  B. 4.48 L  C. 6.74 L  D. 8.96 L

Solution:
T = 273 + 273 = 546 K  R = 8.314 \(n(N_2) = \frac{1.4}{28} = 0.05\ \text{mol}\)
\(n(Ar) = \frac{4.0}{40} = 0.10\ \text{mol}\)
Total n = 0.15 mol

Use PV = nRT:
\[ V = \frac{nRT}{P} \] \[ V = \frac{0.15 \times 8.314 \times 546}{1.01 \times 10^5} \] \[ V ≈ 0.672\ \text{m}^3 = 6.72\ \text{L} \] Closest: 6.74 L
Answer: C. 6.74 L

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26. For a certain volume of gas under constant pressure, when temperature increases by 1 °C:
A. The volume remains unchanged
B. Volume reduces by \(\frac{1}{273}\)
C. Volume increases by \(\frac{1}{273}\)
D. Gas increases by \(\frac{1}{273}\) of its volume at 0 °C

Solution:
Charles’ Law: \( V \propto T \) (in K)
So for every 1 °C rise (i.e. 1 K), volume increases by \( \frac{1}{273} \) of volume at 0 °C.
Answer: D. Gas increases by \(\frac{1}{273}\) of its volume at 0 °C

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27. \( \text{MH}_2(g) + 2H_2O(l) → M(OH)_2(s) + 2H_2(g) \)
0.42 g of compound MH₂ reacted with water and 448 mL of dry hydrogen gas was liberated at STP. Calculate the relative atomic mass of M.
A. 24.3  B. 32  C. 40  D. 42

Solution:
At STP, 1 mol gas = 22400 mL
\[ n(H_2) = \frac{448}{22400} = 0.02\ \text{mol} \] From equation: 1 mol MH₂ → 2 mol H₂
So \( n(MH_2) = \frac{0.02}{2} = 0.01 \) mol \[ M(MH_2) = \frac{0.42}{0.01} = 42 \] \[ M = 42 - 2 = 40 \] Answer: C. 40

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28. A, B and C are three gases in a container. The ratio of the number of molecules is A:B:C = 1:2:3. If the partial pressure of A is \(3 \times 10^4\ \text{Pa}\), which is correct?
I. Partial pressure of C is \(9 \times 10^4\ \text{Pa}\)
II. Total pressure is \(1.8 \times 10^5\ \text{Pa}\)
III. Smaller number of molecules → smaller partial pressure
IV. Mass of gas A is \(\frac{1}{67}\) of the total mass of gases
A. I, II  B. I, IV  C. III, IV  D. I, II, III

Solution:
Pressure ∝ number of molecules
→ C has 3× molecules → pressure = 3× that of A
→ \(P_C = 9 \times 10^4\ \text{Pa}\), \(P_B = 6 \times 10^4\ \text{Pa}\), total = \(1.8 \times 10^5\)

I ✅ II ✅ III ✅ IV ❌ (not enough info on molar mass)
Answer: D. I, II, III

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29. A gas in a 3 L container has an original pressure of 2 kPa. It is allowed to flow into a 17 L container. What is the final pressure of the gas?
A. 0.03 kPa  B. 0.3 kPa  C. 3 kPa  D. 30 kPa

Solution:
Use \(P_1V_1 = P_2V_2\)
\[ 2 \times 3 = P_2 \times (3 + 17) \] \[ 2 \ times 3 = P_2 \times 20 \] \[ P_2 = \frac{6}{20} = 0.3\ \text{kPa} \] Answer: B. 0.3 kPa

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30. If the rate of diffusion for a monoatomic gas is one-third that of hydrogen gas, what is the relative atomic mass of the gas?
A. 4  B. 14  C. 18  D. 20

Solution:
Use Graham's Law of diffusion:
\[ \frac{r_{\text{gas}}}{r_{\text{H}_2}} = \sqrt{ \frac{M_{\text{H}_2}}{M_{\text{gas}}} } \] Given \( \frac{r_{\text{gas}}}{r_{\text{H}_2}} = \frac{1}{3} \), and \( M_{\text{H}_2} = 2 \):
\[ \frac{1}{3} = \sqrt{ \frac{2}{M} } \] \[ \left( \frac{1}{3} \right)^2 = \frac{2}{M} \] \[ \frac{1}{9} = \frac{2}{M} \] \[ M = 18 \] Answer: C. 18

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31. What does 2.2 g of carbon dioxide gas contain?
A. \( 3.0 \times 10^{22} \) atoms
B. 0.5 mol CO₂ molecules
C. Same number of molecules as 2.2 g propane (C₃H₈) gas
D. Half the number of hydrogen atoms in 2.24 L hydrogen gas at STP

Solution:
M(CO₂) = 44 → moles = \( \frac{2.2}{44} = 0.05\ \text{mol} \)
A. Atoms = \( 0.05 \times 3 \times 6.022 \times 10^{23} ≈ 9 \times 10^{22} \) ❌
B. 0.5 mol? No, it’s 0.05 mol ❌
C. Compare: M(C₃H₈) = 44 → 2.2 g = 0.05 mol → same mol → ✅
D. 2.24 L H₂ = 0.1 mol → 0.2 mol atoms → CO₂ has 0.1 mol O atoms → ❌
Answer: C. Same number of molecules as 2.2 g propane (C₃H₈) gas

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32. A hydrogen gas thermometer shows 100 cm³ at 0 °C. It reads 87.2 cm³ in boiling chlorine at same pressure. What is chlorine's boiling point?
A. -35 °C  B. 40 °C  C. 238 °C  D. 313 °C

Solution:
Use Charles' Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \] \[ \frac{100}{273} = \frac{87.2}{T} \] \[ T = \frac{87.2 \times 273}{100} \] \[ T ≈ 238.26\ \text{K} \] \[ T = 238.26 - 273 = -34.7\ ^\circ\text{C} \] Answer: A. -35 °C

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33. 3.172 g of metal M reacts at 20 °C and 1.01×10⁵ Pa with 0.6015 L oxygen to form MO. Oxygen gas density = 1.328 g/L at same conditions. What is M's relative atomic mass?
A. 127  B. 63.5  C. 31.75  D. 23.07

Solution:
Mass of O₂ = \(0.6015 \times 1.328 = 0.798\ \text{g}\)
Moles of O₂ = \( \frac{0.798}{32} = 0.02494 \) mol
From MO: 1 mol O₂ reacts with 2 mol M →
Moles of M = \( 2 \times 0.02494 = 0.04988 \)
Molar mass = \( \frac{3.172}{0.04988} ≈ 63.6 \)
Answer: B. 63.5

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34. 4.0 L methane at \(2.02 \times 10^5\) Pa, 12.5 L nitrogen at \(3.5 \times 10^5\) Pa, and 1.5 L helium at \(1.01 \times 10^5\) Pa are injected into a 10 L vessel. What is the final pressure?
A. \( 5.33 \times 10^6 \) Pa  B. \( 6.53 \times 10^5 \) Pa  C. \( 5.33 \times 10^5 \) Pa  D. \( 6.53 \times 10^4 \) Pa

Solution:
Use Boyle’s Law: \( P_1V_1 = P_2V_2 \)
\[ P = \frac{(4.0)(2.02) + (12.5)(3.5) + (1.5)(1.01)}{10} \times 10^5 \] \[ P = 5.335 \times 10^5 \] Answer: C. \( 5.33 \times 10^5 \) Pa

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35. At 27 °C and pressure 308 kPa, a gas has volume 0.162 L. Its relative molecular mass is 60. What is the mass?
A. 1.2g  B. 2.4g  C. 12g  D. 24g

Solution:
Use \( PV = nRT \)
T = 300 K  R = 8.314
\[ n = \frac{PV}{RT} \] \[ n = \frac{308000 \times 0.000162}{8.314 \times 300} \] \[ n ≈ 0.020\ \text{mol} \] Mass = \( 0.02 \times 60 = 1.2g \)
Answer: A. 1.2g

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36. There are two separated flasks containing 5 L of gas X and 10 L of gas Y respectively. The pressure of gas X is 9 Pa while the pressure of gas Y is 3 Pa. Given that gas X and gas Y do not react with each other, if the valve is opened to allow mixing, what is the total pressure of the mixture?
A. 2 Pa  B. 6 Pa  C. 5 Pa  D. 12 Pa

Solution:
Use total pressure = \( \frac{P_1V_1 + P_2V_2}{V_1 + V_2} \)
\[ P = \frac{9 \times 5 + 3 \times 10}{5 + 10} \] \[ P = \frac{45 + 30}{15} = \frac{75}{15} \] \[ P = 5\ \text{Pa} \] Answer: C. 5 Pa

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37. Given \(P_N\) is the vapour pressure of liquid nitrogen at −T °C. Find the partial vapour pressure of nitrogen for a mixture of liquid nitrogen and oxygen at −T °C, if the mole fraction of nitrogen is \(x\).
A. \(xP_N\)
B. \((1 - x)P_N\)
C. \(\frac{x}{1 - x}P_N\)
D. \(\frac{1 - x}{x}P_N\)

Solution:
Raoult’s Law: partial vapour pressure = mole fraction × vapour pressure of pure solvent
So for nitrogen:
\[ P_{\text{N, partial}} = xP_N \] Answer: A. \(xP_N\)

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38. When 1 mL of ice is heated to 323 °C at 1 atm (101 kPa), what is the volume of steam produced? (Density of ice = 1.0 g/mL)
A. 0.273 L  B. 1.33 L  C. 2.73 L  D. 48.0 L

Solution:
Mass of ice = 1 g → Moles of H₂O = \( \frac{1}{18} ≈ 0.0556 \) mol
Temperature = 323 °C = 596 K
Use ideal gas equation:
\[ V = \frac{nRT}{P} \] \[ V = \frac{0.0556 \times 8.314 \times 596}{101000} \] \[ V ≈ 0.0273\ \text{m}^3 = 27.3\ \text{L} \] Answer: D. 48.0 L

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39. 2.0 mol argon gas and 2.0 mol hydrogen gas are mixed in a vessel, and effuse through a small pinhole. If 0.2 mol hydrogen remains, how much argon is left?
A. 0.2 mol  B. 0.4 mol  C. 1.6 mol  D. 1.8 mol

Solution:
Rate of effusion ∝ \( \frac{1}{\sqrt{M}} \)
Hydrogen escapes faster than argon
Let remaining H₂ = 0.2 mol → 1.8 mol escaped \[ \frac{\text{Rate of Ar}}{\text{Rate of H}_2} = \sqrt{\frac{2}{40}} \] \[ = \frac{1}{\sqrt{20}} \approx \frac{1}{4.47} \] So if H₂ lost 1.8 mol, then Ar lost ≈ \( \frac{1.8}{4.47} ≈ 0.4 \) mol
Remaining Ar = 2.0 − 0.4 = 1.6 mol
Answer: C. 1.6 mol

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40. Which condition is closest to ideal gas behaviour?
A. CO: 0 °C, 50 atm
B. CO: 350 K, 0.1 atm
C. He: 0 °C, 50 atm
D. He: 350 K, 0.1 atm

Solution:
Ideal gas favours: low pressure, high temperature, and non-polar small molecules.
→ Helium is monatomic and inert
→ D: High temp, low pressure
Answer: D. He: 350 K, 0.1 atm

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41. 40 mL CO₂ and 20 mL O₂ are in a closed container at STP. Then 150 mL H₂ is injected later. Which deduction is NOT true?
A. Total pressure increases
B. Partial pressure of CO₂ remains constant
C. Partial pressure of O₂ increases
D. Partial pressure of H₂ is the highest

Solution:
A ✅ more gas → more pressure
B ✅ CO₂ amount unchanged
C ❌ O₂ unchanged → pressure unchanged
D ✅ 150 mL is largest volume
Answer: C. Partial pressure of O₂ increases

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42. Flask X: 2 L argon at 4 kPa. Flask Y: 3 L helium at 2 kPa. Connected at constant temp. What’s final total pressure?
A. 1.2 kPa  B. 2.8 kPa  C. 4.0 kPa  D. 6.0 kPa

Solution:
\[ P = \frac{P_1V_1 + P_2V_2}{V_1 + V_2} \] \[ P = \frac{4 \times 2 + 2 \times 3}{2 + 3} \] \[ P = \frac{8 + 6}{5} = 2.8\ \text{kPa} \] Answer: B. 2.8 kPa

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43. Under standard conditions, the volume of 1.68 g of an unknown gas was measured to be 448 mL. What is the molar mass of this gas?
A. 84 g mol⁻¹  B. 34 g mol⁻¹  C. 0.084 g mol⁻¹  D. 0.034 g mol⁻¹

Solution:
At STP, 1 mole of gas = 22.4 L = 22400 mL
Moles = \( \frac{448}{22400} = 0.02 \) mol
Molar mass = \( \frac{1.68}{0.02} = 84 \)
Answer: A. 84 g mol⁻¹

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44. At specific pressure and temperature, 4 L of a gas mixture contains 66 g CO₂ and \(x\) g O₂. After passing through concentrated NaOH, the volume becomes 2.5 L. What is \(x\)?
A. 40 g  B. 60 g  C. 80 g  D. 100 g

Solution:
NaOH removes CO₂ → CO₂ volume = 4 L − 2.5 L = 1.5 L
Moles of CO₂ = \( \frac{66}{44} = 1.5 \) mol → 1.5 L at same condition
So O₂ is 2.5 L → same volume as 2.5 mol if equal conditions
Mass = \( 2.5 \times 32 = 80 \) g
Answer: C. 80 g

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Will add more soon, perhaps